Tüm alıştırma soruları

2131 soru

Soru 1781Soru

For all positive real numbers aa and bb, the custom operation \diamondsuit is defined by ab=a2+b2aba \diamondsuit b = \frac{a^2 + b^2}{ab}. The function ff is defined for all x>0x > 0 by f(x)=x4f(x) = x \diamondsuit 4. If f(x)=2.5f(x) = 2.5, what is the value of xx that is greater than 44?

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Cevap: 8

Cevap

The value of xx greater than 44 is 88.
Applying the custom operator gives f(x)=x2+164xf(x) = \frac{x^2 + 16}{4x}. Setting this equal to 2.52.5 yields x2+164x=52\frac{x^2 + 16}{4x} = \frac{5}{2}, which simplifies to x210x+16=0x^2 - 10x + 16 = 0. The roots are x=2x = 2 and x=8x = 8. Since xx must be greater than 44, the only valid answer is 88.

Adım Adım Çözüm

1
Substitute a=xa = x and b=4b = 4 into the custom operation definition ab=a2+b2aba \diamondsuit b = \frac{a^2 + b^2}{ab}.
f(x)=x2+164xf(x) = \frac{x^2 + 16}{4x}
This establishes the explicit algebraic rule for the function f(x)f(x).
2
Set f(x)f(x) equal to 2.52.5 and clear the fraction.
x2+164x=2.5    x2+16=10x\frac{x^2 + 16}{4x} = 2.5 \implies x^2 + 16 = 10x
Multiplying both sides by 4x4x converts the rational equation into a standard polynomial equation.
3
Rearrange into standard quadratic form and solve by factoring.
x210x+16=0    (x2)(x8)=0    x=2x^2 - 10x + 16 = 0 \implies (x - 2)(x - 8) = 0 \implies x = 2 or x=8x = 8
Factoring determines all potential positive real solutions for xx.
4
Select the solution satisfying the constraint x>4x > 4.
x=8x = 8
The question explicitly specifies that xx must be greater than 44, eliminating x=2x = 2.

Anahtar Kavram

Evaluating custom binary operations and solving algebraic function equations involving quadratic constraints.
Soru 1782Soru

A container holds nn spheres, exactly 5 of which are blue and the remaining n5n - 5 are green. Two spheres are drawn at random from the container one after another without replacement. If the probability that at least one of the selected spheres is green is 1415\frac{14}{15}, what is the total number of spheres nn in the container?

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Cevap: 25

Cevap

The total number of spheres nn in the container is 25.
The probability of at least one green sphere is complementary to drawing zero green spheres (meaning both spheres drawn are blue). Subtracting 1415\frac{14}{15} from 1 yields P(both blue)=115P(\text{both blue}) = \frac{1}{15}. Since the selection is without replacement, P(both blue)=5n×4n1=20n(n1)P(\text{both blue}) = \frac{5}{n} \times \frac{4}{n-1} = \frac{20}{n(n-1)}. Equating this to 115\frac{1}{15} gives n(n1)=300n(n-1) = 300. Solving the quadratic equation n2n300=0n^2 - n - 300 = 0 gives n=25n = 25, which correctly represents the total number of spheres.

Adım Adım Çözüm

1
Use the complement rule to determine the probability that both selected spheres are blue.
P(both blue)=1P(at least one green)=11415=115P(\text{both blue}) = 1 - P(\text{at least one green}) = 1 - \frac{14}{15} = \frac{1}{15}.
The event that at least one sphere is green is the complement of the event that both drawn spheres are blue.
2
Set up the joint probability equation for drawing two blue spheres sequentially without replacement.
P(both blue)=5n×4n1=20n(n1)P(\text{both blue}) = \frac{5}{n} \times \frac{4}{n - 1} = \frac{20}{n(n - 1)}.
There are 5 blue spheres initially out of nn. After drawing one blue sphere, 4 blue spheres remain out of n1n - 1 total spheres.
3
Equate the expressions and solve for nn.
\begin{aligned} \frac{20}{n(n - 1)} &= \frac{1}{15} \\ n(n - 1) &= 300 \\ n^2 - n - 300 &= 0 \\ (n - 25)(n + 12) &= 0 \end{aligned}
Cross-multiplying gives a quadratic equation in terms of nn.
4
Select the valid positive integer solution for nn.
n=25n = 25 (since n>0n > 0).
The total number of spheres must be a positive integer.

Anahtar Kavram

Probability of Complementary Events and Dependent Sequential Events
Tahmini Süre:2m 0s
Soru 1783Soru

A courier service calculates the shipping cost for a package based on its weight ww, in pounds. For packages weighing up to 2020 pounds, the cost is a flat fee of $12.00\$12.00 plus $2.50\$2.50 per pound. For packages weighing more than 2020 pounds, the cost is a flat fee of $20.00\$20.00 plus $2.00\$2.00 per pound. If the total shipping cost for a package was strictly greater than $45.00\$45.00 and at most $70.00\$70.00, which of the following could be the weight of the package, in pounds? Select all such weights.

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Cevap: 1515 pounds; 2020 pounds; 2424 pounds

Cevap

The valid weights of the package are 15 pounds, 20 pounds, and 24 pounds.
The weight of the package must result in a total cost C(w)C(w) such that $45.00<C(w)$70.00\$45.00 < C(w) \le \$70.00. Evaluating the options:
- For 15 pounds: 12+2.50(15)=$49.5012 + 2.50(15) = \$49.50, which falls within the range.
- For 20 pounds: 12+2.50(20)=$62.0012 + 2.50(20) = \$62.00, which falls within the range.
- For 24 pounds: 20+2.00(24)=$68.0020 + 2.00(24) = \$68.00, which falls within the range.

Adım Adım Çözüm

1
Formulate the cost function C(w)C(w) as a piecewise model.
C(w)=12+2.50wC(w) = 12 + 2.50w for 0<w200 < w \le 20, and C(w)=20+2.00wC(w) = 20 + 2.00w for w>20w > 20.
The cost structure changes depending on whether the weight exceeds 2020 pounds.
2
Set up and solve the inequality 45<C(w)7045 < C(w) \le 70 for packages up to 2020 pounds.
45<12+2.50w70    33<2.50w58    13.2<w23.245 < 12 + 2.50w \le 70 \implies 33 < 2.50w \le 58 \implies 13.2 < w \le 23.2. Restricting to w20w \le 20 gives 13.2<w2013.2 < w \le 20.
This determines the valid weight interval for packages billed under the first tier.
3
Set up and solve the inequality 45<C(w)7045 < C(w) \le 70 for packages over 2020 pounds.
45<20+2.00w70    25<2.00w50    12.5<w2545 < 20 + 2.00w \le 70 \implies 25 < 2.00w \le 50 \implies 12.5 < w \le 25. Restricting to w>20w > 20 gives 20<w2520 < w \le 25.
This determines the valid weight interval for packages billed under the second tier.
4
Combine the valid intervals and evaluate each option.
The overall valid weight interval is 13.2<w2513.2 < w \le 25. Testing the values: 12 pounds is outside the range; 15 pounds, 20 pounds, and 24 pounds are within the range; 26 pounds is outside the range.
Any weight strictly greater than 13.213.2 pounds and up to 2525 pounds produces a cost between $45.00\$45.00 and $70.00\$70.00.

Anahtar Kavram

Algebraic Modeling of Piecewise Functions and Compound Inequalities
Soru 1784Soru

A short-pulse laser emits a single pulse lasting 8.4×1098.4 \times 10^{-9} seconds. A high-speed optical sensor completes one measurement cycle every 1.4×10111.4 \times 10^{-11} seconds. How many measurement cycles does the sensor complete during the duration of a single laser pulse?

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Cevap: 600

Cevap

600
Dividing the pulse duration (8.4×1098.4 \times 10^{-9} seconds) by the sensor cycle time (1.4×10111.4 \times 10^{-11} seconds) yields 8.41.4×109(11)=6×102=600\frac{8.4}{1.4} \times 10^{-9 - (-11)} = 6 \times 10^2 = 600 complete cycles.

Adım Adım Çözüm

1
Set up the division expression for the total number of cycles.
\frac{8.4 \times 10^{-9}\text{ seconds}}{1.4 \times 10^{-11}\text{ seconds}}
To find how many cycle intervals fit within the total pulse duration.
2
Divide the decimal coefficients.
8.41.4=6\frac{8.4}{1.4} = 6
Separating the numerical coefficients from the powers of ten.
3
Apply exponent rules to divide powers of 10.
10^{-9 - (-11)} = 10^{-9 + 11} = 10^2 = 100
Dividing powers with the same base requires subtracting the denominator exponent from the numerator exponent.
4
Combine results to find total cycles.
6×100=6006 \times 100 = 600
Multiplying coefficient quotient by the simplified power of ten.

Anahtar Kavram

Division of Numbers in Scientific Notation and Exponent Rules
Soru 1785Soru

A dataset SS consists of 7 distinct positive integers. The arithmetic mean of the numbers in SS is 20, and the median is 18. If the largest integer in SS is 35, what is the maximum possible value for the second-largest integer in SS?

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Cevap: 34

Cevap

34
To find the maximum possible value of the second-largest integer, we arrange the 7 distinct positive integers in ascending order: x1<x2<x3<x4<x5<x6<x7x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7. The median x4=18x_4 = 18 and the largest element x7=35x_7 = 35. The total sum of all 7 elements is 7×20=1407 \times 20 = 140. Since x6x_6 must be strictly less than x7=35x_7 = 35, the maximum potential integer value for x6x_6 is 34. If x6=34x_6 = 34, the remaining four elements (x1,x2,x3,x5x_1, x_2, x_3, x_5) must sum to 8734=5387 - 34 = 53. Choosing x5=19x_5 = 19 (the smallest integer greater than 18) leaves a sum of 34 for x1+x2+x3x_1 + x_2 + x_3, which can be satisfied by distinct positive integers such as 1, 16, and 17. Thus, 34 is achievable.

Adım Adım Çözüm

1
Calculate the total sum of the 7 integers in the dataset.
Total sum = 7×20=1407 \times 20 = 140.
The sum of a dataset equals the number of elements multiplied by its arithmetic mean.
2
Identify the positions of known elements when the dataset is ordered in ascending order x1<x2<x3<x4<x5<x6<x7x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7.
Median x4=18x_4 = 18 and largest element x7=35x_7 = 35.
For a 7-element dataset, the median is the 4th element.
3
Calculate the combined sum of the remaining five unknown elements.
x1+x2+x3+x5+x6=140(18+35)=87x_1 + x_2 + x_3 + x_5 + x_6 = 140 - (18 + 35) = 87.
Subtracting the median and the largest element from the total sum gives the sum of the remaining five elements.
4
Determine the theoretical upper bound for the second-largest integer x6x_6.
Since x6<x7=35x_6 < x_7 = 35 and all integers are distinct, x634x_6 \le 34.
The second-largest integer must be strictly less than the largest integer.
5
Verify if x6=34x_6 = 34 can produce a valid dataset of distinct positive integers.
If x6=34x_6 = 34, then x1+x2+x3+x5=8734=53x_1 + x_2 + x_3 + x_5 = 87 - 34 = 53. Setting x5=19x_5 = 19, x3=17x_3 = 17, x2=16x_2 = 16, and x1=1x_1 = 1 gives 1+16+17+19=531 + 16 + 17 + 19 = 53, forming the valid set {1,16,17,18,19,34,35}\{1, 16, 17, 18, 19, 34, 35\}.
Since a valid set of distinct positive integers exists satisfying all constraints, 34 is the maximum possible value.

Anahtar Kavram

Measures of Central Tendency and Data Constraints
Soru 1786Soru

An automated risk-management system uses three independent algorithms—Algorithm X, Algorithm Y, and Algorithm Z—to detect fraudulent transactions. The probability that Algorithm X detects a given fraudulent transaction is 35\frac{3}{5}, the probability that Algorithm Y detects it is 23\frac{2}{3}, and the probability that Algorithm Z detects it is 34\frac{3}{4}. If a fraudulent transaction occurs, what is the probability that it will be detected by at least two of these three algorithms?

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Cevap: 0.75

Cevap

The probability that the transaction is detected by at least two of the three algorithms is 0.75 (or 3/4).
Because the algorithms operate independently, the event 'at least two algorithms detect the transaction' consists of four mutually exclusive outcomes: exactly X and Y detect (probability 6/60 = 0.10), exactly X and Z detect (probability 9/60 = 0.15), exactly Y and Z detect (probability 12/60 = 0.20), and all three detect (probability 18/60 = 0.30). Summing these four probabilities gives 0.10 + 0.15 + 0.20 + 0.30 = 0.75.

Adım Adım Çözüm

1
Determine the complementary probabilities of non-detection for each algorithm.
P(X does not detect) = 2/5, P(Y does not detect) = 1/3, and P(Z does not detect) = 1/4.
The probability of an event's complement is 1 minus the probability of the event.
2
Calculate the probability for each scenario where exactly two algorithms detect the transaction.
P(X and Y only) = 6/60, P(X and Z only) = 9/60, P(Y and Z only) = 12/60.
Since the algorithms operate independently, joint probabilities are calculated by multiplying individual probabilities.
3
Calculate the probability that all three algorithms detect the transaction.
P(X and Y and Z) = 18/60.
Multiplying the individual detection probabilities of all three independent algorithms.
4
Sum the probabilities of all qualifying mutually exclusive outcomes.
(6/60) + (9/60) + (12/60) + (18/60) = 45/60 = 0.75.
The events representing different combinations of detections are mutually exclusive, so their probabilities add directly.

Anahtar Kavram

Independent Events and Addition Rule for Mutually Exclusive Outcomes
Soru 1787Soru

If pp, p+2p + 2, and p+4p + 4 are all prime numbers, what is the value of p2+5p^2 + 5?

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Cevap: 14

Cevap

The value of p2+5p^2 + 5 is 14.
For any three consecutive odd integers pp, p+2p + 2, and p+4p + 4, exactly one of them must be a multiple of 3. Because all three expressions represent prime numbers, the term divisible by 3 must be equal to 3 (since 3 is the only prime divisible by 3). Setting p=3p = 3 gives p+2=5p + 2 = 5 and p+4=7p + 4 = 7, both of which are prime. Any choice of p>3p > 3 forces either p+2p + 2 or p+4p + 4 to be a multiple of 3 greater than 3, making it composite. Thus p=3p = 3 is uniquely determined, and evaluating p2+5p^2 + 5 gives 32+5=143^2 + 5 = 14.

Adım Adım Çözüm

1
Analyze the possible remainders when the prime pp is divided by 3.
Any positive integer pp can be written in one of three forms: 3k3k, 3k+13k + 1, or 3k+23k + 2 for some integer kk.
Dividing any integer by 3 leaves a remainder of 0, 1, or 2.
2
Test each remainder case for the expressions pp, p+2p + 2, and p+4p + 4.
If p=3k+1p = 3k + 1, then p+2=3k+3=3(k+1)p + 2 = 3k + 3 = 3(k + 1), which is divisible by 3. If p=3k+2p = 3k + 2, then p+4=3k+6=3(k+2)p + 4 = 3k + 6 = 3(k + 2), which is divisible by 3.
Among any three consecutive odd numbers of the form p,p+2,p+4p, p+2, p+4, exactly one of them must be a multiple of 3.
3
Deduce the unique value of pp.
The only way all three numbers pp, p+2p + 2, and p+4p + 4 can be prime is if the one divisible by 3 is equal to 3 itself, which forces p=3p = 3.
The only prime number divisible by 3 is 3 itself; any larger multiple of 3 is composite.
4
Evaluate the target expression p2+5p^2 + 5 using p=3p = 3.
32+5=9+5=143^2 + 5 = 9 + 5 = 14.
Substitute the uniquely determined value p=3p = 3 into the given algebraic expression.

Anahtar Kavram

Divisibility properties of consecutive odd integers and prime number definitions
Tahmini Süre:1m 30s
Soru 1788Soru

A logistics company operates two delivery vans, Van A and Van B. Van A consumes fuel at a rate of 11 gallon for every 1515 miles driven, and Van B consumes fuel at a rate of 11 gallon for every 2525 miles driven. On a specific trip, the two vans were driven a total combined distance of 450450 miles and together consumed exactly 2222 gallons of fuel. What was the total distance, in miles, driven by Van A?

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Cevap: 150 miles

Cevap

The total distance driven by Van A was 150 miles.
The correct answer is 150 miles. By setting up the total fuel consumption equation as dA15+450dA25=22\frac{d_A}{15} + \frac{450 - d_A}{25} = 22, multiplying by the least common multiple 7575 gives 5dA+13503dA=16505d_A + 1350 - 3d_A = 1650, which simplifies to 2dA=3002d_A = 300 and dA=150d_A = 150 miles.

Adım Adım Çözüm

1
Define variables for the distance traveled by each van.
Let dAd_A be the distance driven by Van A. Then the distance driven by Van B is 450dA450 - d_A.
The total combined distance for both vans is 450450 miles.
2
Express the total fuel consumed in terms of dAd_A.
\frac{d_A}{15} + \frac{450 - d_A}{25} = 22
Fuel used equals distance divided by miles per gallon for each vehicle.
3
Clear denominators by multiplying the entire equation by the common denominator 75.
5d_A + 3(450 - d_A) = 1650
Simplifies fractions to solve the linear algebraic equation easily.
4
Solve for dAd_A.
5d_A + 1350 - 3d_A = 1650 \implies 2d_A = 300 \implies d_A = 150
Isolates the target variable dAd_A to find the distance driven by Van A.

Anahtar Kavram

Linear Equations and Rate-Distance Modeling
Soru 1789Soru

In the xyxy-plane, line kk has an xx-intercept of (8,0)(8, 0) and a yy-intercept of (0,6)(0, 6). Line pp is perpendicular to line kk and intersects line kk at its yy-intercept. What is the xx-intercept of line pp?

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Cevap: 92-\frac{9}{2}

Cevap

92-\frac{9}{2}
The line kk passes through (8,0)(8, 0) and (0,6)(0, 6), giving a slope of mk=6008=34m_k = \frac{6 - 0}{0 - 8} = -\frac{3}{4}. A line perpendicular to kk must have a slope equal to the negative reciprocal of 34-\frac{3}{4}, which is 43\frac{4}{3}. Because line pp intersects line kk at (0,6)(0, 6), its yy-intercept is also 66, making its equation y=43x+6y = \frac{4}{3}x + 6. Setting y=0y = 0 gives 0=43x+60 = \frac{4}{3}x + 6, which solves to x=92x = -\frac{9}{2}.

Adım Adım Çözüm

1
Calculate the slope of line kk
Slope of line kk is mk=6008=68=34m_k = \frac{6 - 0}{0 - 8} = -\frac{6}{8} = -\frac{3}{4}
The slope formula is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} using given intercepts (8,0)(8, 0) and (0,6)(0, 6).
2
Determine the slope of line pp
Slope of line pp is mp=1mk=43m_p = -\frac{1}{m_k} = \frac{4}{3}
Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.
3
Write the equation of line pp
Equation of line pp is y=43x+6y = \frac{4}{3}x + 6
Line pp intersects line kk at its yy-intercept (0,6)(0, 6), so line pp has a yy-intercept of 66.
4
Find the xx-intercept of line pp
x=92x = -\frac{9}{2} (or 4.5-4.5)
Set y=0y = 0 in the line equation: 0=43x+6    43x=6    x=634=920 = \frac{4}{3}x + 6 \implies \frac{4}{3}x = -6 \implies x = -6 \cdot \frac{3}{4} = -\frac{9}{2}.

Anahtar Kavram

Perpendicular Slopes and Line Intercepts
Tahmini Süre:1m 30s
Soru 1790Soru

In triangle ABCABC, the measure of angle AA is 4545^\circ and the measure of angle BB is 105105^\circ. The length of side ABAB is 66 units. Which of the following statements must be true? Select all that apply.

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Cevap: The length of the altitude from vertex BB to side ACAC is 323\sqrt{2} units.; The length of side BCBC is 626\sqrt{2} units.; The area of triangle ABCABC is 9+939 + 9\sqrt{3} square units.

Cevap

The correct statements are: the length of the altitude from vertex BB to side ACAC is 323\sqrt{2} units, the length of side BCBC is 626\sqrt{2} units, and the area of triangle ABCABC is 9+939 + 9\sqrt{3} square units.
By drawing altitude BDBD perpendicular to ACAC, triangle ABCABC decomposes into two special right triangles. In the 45459045^\circ-45^\circ-90^\circ triangle ABDABD, the hypotenuse is 66, yielding leg lengths BD=AD=32BD = AD = 3\sqrt{2}. In the 30609030^\circ-60^\circ-90^\circ triangle BCDBCD, side BD=32BD = 3\sqrt{2} is opposite the 3030^\circ angle, making hypotenuse BC=2×32=62BC = 2 \times 3\sqrt{2} = 6\sqrt{2} and long leg CD=36CD = 3\sqrt{6}. Combining ADAD and CDCD gives base AC=32+36AC = 3\sqrt{2} + 3\sqrt{6}, leading to an area of 12(32+36)(32)=9+93\frac{1}{2}(3\sqrt{2} + 3\sqrt{6})(3\sqrt{2}) = 9 + 9\sqrt{3}.

Adım Adım Çözüm

1
Determine the third angle of triangle ABCABC
C=180(45+105)=30\angle C = 180^\circ - (45^\circ + 105^\circ) = 30^\circ.
The sum of interior angles in any triangle is 180180^\circ.
2
Drop an altitude BDBD perpendicular to side ACAC
Altitude BDBD splits triangle ABCABC into two right triangles: ABD\triangle ABD (45459045^\circ-45^\circ-90^\circ) and BCD\triangle BCD (30609030^\circ-60^\circ-90^\circ).
In ABD\triangle ABD, A=45\angle A = 45^\circ and ADB=90\angle ADB = 90^\circ, leaving ABD=45\angle ABD = 45^\circ. In BCD\triangle BCD, CBD=10545=60\angle CBD = 105^\circ - 45^\circ = 60^\circ and C=30\angle C = 30^\circ.
3
Calculate side lengths in 45459045^\circ-45^\circ-90^\circ triangle ABDABD
AD=BD=AB2=62=32AD = BD = \frac{AB}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} units.
The ratio of sides in a 45459045^\circ-45^\circ-90^\circ triangle is 1:1:21:1:\sqrt{2}.
4
Calculate side lengths in 30609030^\circ-60^\circ-90^\circ triangle BCDBCD
Hypotenuse BC=2×BD=62BC = 2 \times BD = 6\sqrt{2} units, and long leg CD=BD×3=32×3=36CD = BD \times \sqrt{3} = 3\sqrt{2} \times \sqrt{3} = 3\sqrt{6} units.
The ratio of sides opposite 30:60:9030^\circ:60^\circ:90^\circ is 1:3:21:\sqrt{3}:2.
5
Calculate total base ACAC and area of triangle ABCABC
AC=AD+CD=32+36AC = AD + CD = 3\sqrt{2} + 3\sqrt{6} units. Area = 12×base×height=12(32+36)(32)=12(18+183)=9+93\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (3\sqrt{2} + 3\sqrt{6})(3\sqrt{2}) = \frac{1}{2} (18 + 18\sqrt{3}) = 9 + 9\sqrt{3} square units.
Standard formula for triangle area is 12bh\frac{1}{2} b h.

Anahtar Kavram

Decomposing an oblique triangle with 4545^\circ and 3030^\circ angles into 45459045^\circ-45^\circ-90^\circ and 30609030^\circ-60^\circ-90^\circ special right triangles.
Tahmini Süre:2m 0s
Soru 1791Soru

A university department tracked the number of research articles published by its 8 faculty members over a five-year period. The numbers of publications for 7 of the faculty members were 4,7,9,12,15,18,4, 7, 9, 12, 15, 18, and 2323. If the arithmetic mean of the number of publications for all 8 faculty members is equal to 1.251.25 times their median, what is the number of publications for the 8th faculty member?

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Cevap: 47

Cevap

47
The sum of the 7 known values is 8888, making the total sum 88+x88 + x and the arithmetic mean 88+x8\frac{88 + x}{8}. Assuming x15x \ge 15, the 4th and 5th numbers when sorted are 1212 and 1515, yielding a median of 13.513.5. Setting the mean equal to 1.25×13.5=16.8751.25 \times 13.5 = 16.875 gives 88+x8=16.875\frac{88 + x}{8} = 16.875, which simplifies to 88+x=13588 + x = 135 and yields x=47x = 47.

Adım Adım Çözüm

1
Calculate the sum of the 7 known data points and express the mean in terms of the unknown 8th value xx.
Sum of 7 known values = 4+7+9+12+15+18+23=884 + 7 + 9 + 12 + 15 + 18 + 23 = 88. Total mean = 88+x8\frac{88 + x}{8}.
The arithmetic mean of nn values is the sum of all values divided by nn.
2
Analyze the position of xx in sorted order to determine the median.
Assuming x15x \ge 15, the 4th and 5th values in ascending order are 1212 and 1515, giving a median of 12+152=13.5\frac{12 + 15}{2} = 13.5.
For an even number of data points (88), the median is the average of the 4th and 5th terms in sorted order.
3
Formulate and solve the equation linking the mean and median.
88+x8=1.25×13.5=16.875    88+x=135    x=47\frac{88 + x}{8} = 1.25 \times 13.5 = 16.875 \implies 88 + x = 135 \implies x = 47.
The problem specifies that the mean is equal to 1.251.25 times the median.

Anahtar Kavram

Calculating mean and median of a dataset containing an unknown value.
Tahmini Süre:1m 30s
Soru 1792Soru

A laboratory technician has 1212 liters of a solution containing 15%15\% salt by weight. The technician wants to increase the salt concentration to 25%25\% by adding a second solution that contains 40%40\% salt by weight. How many liters of the 40%40\% solution must be added?

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Cevap: 8

Cevap

8 liters of the 40% solution must be added.
The total amount of salt contributed by the two solutions is 0.15(12)+0.40x=1.8+0.40x0.15(12) + 0.40x = 1.8 + 0.40x. The final mixture volume is (12+x)(12 + x) liters, and its concentration must be 25%25\%. Equating the total salt to 25%25\% of total volume yields 1.8+0.40x=0.25(12+x)1.8 + 0.40x = 0.25(12 + x). Expanding gives 1.8+0.40x=3.0+0.25x1.8 + 0.40x = 3.0 + 0.25x, which simplifies to 0.15x=1.20.15x = 1.2, giving x=8x = 8 liters.

Adım Adım Çözüm

1
Define the unknown variable and calculate the initial solute quantity.
Let xx be the volume of the 40%40\% solution added (in liters). Salt in initial solution = 0.15×12=1.80.15 \times 12 = 1.8 liters.
Establishing the mass balance of the salt solute is necessary to construct the algebraic equation.
2
Set up the algebraic concentration equation.
1.8+0.40x=0.25(12+x)1.8 + 0.40x = 0.25(12 + x)
The combined salt from both solutions must equal 25%25\% of the total combined liquid volume (12+x)(12 + x) liters.
3
Solve the linear equation for xx.
1.8+0.40x=3.0+0.25x    0.15x=1.2    x=81.8 + 0.40x = 3.0 + 0.25x \implies 0.15x = 1.2 \implies x = 8
Isolating xx yields the exact number of liters required.

Anahtar Kavram

Algebraic mixture problems using mass balance equations.
Tahmini Süre:1m 30s
Soru 1793Soru

A committee of 33 members is to be selected at random without replacement from a group of 55 data scientists and 55 software engineers. What is the conditional probability that at least two data scientists are selected in total, given that the first person selected is a data scientist?

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Cevap: 1318\frac{13}{18}

Cevap

The conditional probability is 1318\frac{13}{18}.
The conditional probability is calculated by adjusting the sample space after the first draw. With 1 data scientist already selected, there are 4 data scientists and 5 software engineers left (9 total). Selecting at least 1 more data scientist in the next 2 draws has a probability complementary to selecting 0 more data scientists. The probability of selecting 0 additional data scientists is (52)(92)=1036=518\frac{\binom{5}{2}}{\binom{9}{2}} = \frac{10}{36} = \frac{5}{18}. Therefore, the required conditional probability is 1518=13181 - \frac{5}{18} = \frac{13}{18}.

Adım Adım Çözüm

1
Determine the remaining pool of candidates after the first selection.
Since 1 data scientist is selected first, 4 data scientists and 5 software engineers remain (total of 9 candidates). Two more candidates must be selected.
The conditional statement fixes the first selection, changing the sample space for the remaining 2 selections.
2
Identify the condition required for the target event to occur.
Since 1 data scientist is already selected, having at least two data scientists in total means selecting at least 1 additional data scientist in the remaining 2 draws.
Total data scientists = 1 (first draw) + (number of data scientists in next 2 draws).
3
Calculate the complementary probability (selecting 0 additional data scientists).
The number of ways to pick 2 software engineers from 5 is (52)=10\binom{5}{2} = 10. The total ways to pick 2 people from 9 is (92)=36\binom{9}{2} = 36. The probability of 0 additional data scientists is 1036=518\frac{10}{36} = \frac{5}{18}.
Selecting 0 additional data scientists is equivalent to selecting 2 software engineers from the remaining group.
4
Subtract the complementary probability from 1.
1518=13181 - \frac{5}{18} = \frac{13}{18}.
The sum of the probability of an event and its complement equals 1.

Anahtar Kavram

Conditional Probability without Replacement
Soru 1794Soru

A museum display curator is arranging 66 distinct historical coins—44 silver coins and 22 gold coins—in a single row inside a display case. If the 22 gold coins cannot be placed adjacent to each other, how many different linear arrangements of the 66 coins are possible?

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Cevap: 480480

Cevap

The total number of valid linear arrangements is 480480.
Arranging the 44 distinct silver coins yields 4!=244! = 24 orderings. These 44 coins create 55 available gaps (including both ends). To guarantee the 22 distinct gold coins are not adjacent, each gold coin must occupy a separate gap. The number of ways to assign 22 distinct gold coins to 55 gaps is P(5,2)=5×4=20P(5, 2) = 5 \times 4 = 20. By the Fundamental Counting Principle, the total number of valid linear arrangements is 24×20=48024 \times 20 = 480.

Adım Adım Çözüm

1
Calculate the total number of ways to arrange the unrestricted silver coins.
The 44 distinct silver coins can be arranged in 4!=244! = 24 ways.
The relative positions of the silver coins matter because each coin is distinct.
2
Determine the number of available gap positions created by the silver coins to separate the gold coins.
Arranging 44 silver coins creates 55 possible gaps (one before the first coin, three between adjacent silver coins, and one after the last coin: \_ S \_ S \_ S \_ S \_).
Placing at most one gold coin per gap guarantees that the two gold coins will not be adjacent.
3
Calculate the number of ways to place the 22 distinct gold coins into the 55 available gaps.
The number of ways to place the 22 distinct gold coins into 55 distinct slots is P(5,2)=5×4=20P(5, 2) = 5 \times 4 = 20.
Order matters because the gold coins are distinct objects.
4
Apply the Fundamental Counting Principle to find the total arrangements.
Total arrangements = 24×20=48024 \times 20 = 480.
The arrangement of the silver coins and the placement of the gold coins are independent sequential choices.

Anahtar Kavram

Slot method for non-adjacent arrangements using permutations and the Fundamental Counting Principle
Soru 1795Soru

In the xyxy-plane, line LL has a slope of 34-\frac{3}{4} and passes through the point (4,1)(4, 1). Which of the following statements about line LL must be true? Select all such statements.

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Cevap: Line LL passes through the point (4,7)(-4, 7).; Line LL is perpendicular to the line defined by 4x3y=124x - 3y = 12.; Line LL is parallel to the line defined by 3x+4y=253x + 4y = 25.

Cevap

The true statements are that line L passes through (-4, 7), is perpendicular to 4x - 3y = 12, and is parallel to 3x + 4y = 25.
The correct statements are those identifying that (4,7)(-4, 7) satisfies y=34x+4y = -\frac{3}{4}x + 4, that 4x3y=124x - 3y = 12 has a perpendicular slope of 43\frac{4}{3}, and that 3x+4y=253x + 4y = 25 has an identical parallel slope of 34-\frac{3}{4}.

Adım Adım Çözüm

1
Find the equation of line L using point-slope form.
y1=34(x4)    y=34x+4y - 1 = -\frac{3}{4}(x - 4) \implies y = -\frac{3}{4}x + 4, or in standard form 3x+4y=163x + 4y = 16.
Establishing the linear equation allows systematic verification of points, intercepts, and slope properties.
2
Verify point (4,7)(-4, 7) on line L.
3(4)+4(7)=12+28=163(-4) + 4(7) = -12 + 28 = 16, which satisfies the equation.
Checking if the point coordinates satisfy 3x+4y=163x + 4y = 16 confirms it lies on the line.
3
Determine quadrant coverage.
The yy-intercept is (0,4)(0, 4) and the xx-intercept is (163,0)\left(\frac{16}{3}, 0\right). For x<0x < 0, y>4y > 4 (Quadrant II); for 0<x<1630 < x < \frac{16}{3}, y>0y > 0 (Quadrant I); for x>163x > \frac{16}{3}, y<0y < 0 (Quadrant IV).
Line L passes through Quadrants I, II, and IV only, so it does not enter Quadrant III.
4
Analyze slope relationships for perpendicularity and parallelism.
Line 4x3y=124x - 3y = 12 has slope 43\frac{4}{3} (negative reciprocal of 34-\frac{3}{4}). Line 3x+4y=253x + 4y = 25 has slope 34-\frac{3}{4} (identical slope).
Perpendicular lines have slopes that multiply to 1-1, while parallel lines have equal slopes and different intercepts.

Anahtar Kavram

Linear line equations, parallel/perpendicular slopes, and point/intercept evaluations in coordinate geometry.
Tahmini Süre:1m 30s
Soru 1796Soru

A robot starting at point PP travels due east for 44 meters to point QQ. At point QQ, the robot turns 6060^\circ counterclockwise from its original direction and travels in a straight line for 88 meters, stopping at point RR. What is the straight-line distance, in meters, between point PP and point RR?

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Cevap: 474\sqrt{7}

Cevap

The straight-line distance between point PP and point RR is 474\sqrt{7} meters.
Dropping a vertical perpendicular from RR to the extended line PQPQ forms a 3030^\circ-6060^\circ-9090^\circ right triangle QSRQSR with hypotenuse 88. The base extension QSQS equals 44 and the altitude RSRS equals 434\sqrt{3}. Applying the Pythagorean theorem to the right triangle PSRPSR with legs PS=4+4=8PS = 4 + 4 = 8 and RS=43RS = 4\sqrt{3} gives PR=82+(43)2=64+48=112=47PR = \sqrt{8^2 + (4\sqrt{3})^2} = \sqrt{64 + 48} = \sqrt{112} = 4\sqrt{7} meters.

Adım Adım Çözüm

1
Construct a right triangle by extending segment PQPQ past QQ to a point SS directly below RR, dropping altitude RSPSRS \perp PS.
Triangle QSRQSR is formed with RQS=60\angle RQS = 60^\circ and QSR=90\angle QSR = 90^\circ.
Decomposing the angled path into perpendicular horizontal and vertical components allows the use of special right triangles and the Pythagorean theorem.
2
Use 3030^\circ-6060^\circ-9090^\circ side ratios (1:3:21 : \sqrt{3} : 2) in triangle QSRQSR where hypotenuse QR=8QR = 8.
Horizontal segment QS=8×12=4QS = 8 \times \frac{1}{2} = 4 meters, and vertical altitude RS=8×32=43RS = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3} meters.
The leg opposite the 3030^\circ angle is half the hypotenuse, and the leg opposite the 6060^\circ angle is 3\sqrt{3} times the short leg.
3
Calculate the total horizontal distance PSPS and vertical distance RSRS for the large right triangle PSRPSR.
Total base PS=PQ+QS=4+4=8PS = PQ + QS = 4 + 4 = 8 meters, and height RS=43RS = 4\sqrt{3} meters.
Point PP, point QQ, and point SS are collinear on the horizontal line.
4
Apply the Pythagorean theorem to right triangle PSRPSR to find distance PRPR.
PR=PS2+RS2=82+(43)2=64+48=112=47PR = \sqrt{PS^2 + RS^2} = \sqrt{8^2 + (4\sqrt{3})^2} = \sqrt{64 + 48} = \sqrt{112} = 4\sqrt{7} meters.
The straight-line distance PRPR is the hypotenuse of right triangle PSRPSR.

Anahtar Kavram

Combining 3030^\circ-6060^\circ-9090^\circ special right triangles with the Pythagorean theorem to solve multi-step 2D path/distance problems.
Tahmini Süre:1m 45s
Soru 1797Soru

A healthcare clinic logged the waiting times, in minutes, for 7 patients on Monday: 12,41,24,19,33,28,12, 41, 24, 19, 33, 28, and 4949. On Tuesday, 3 additional patients were logged. The arithmetic mean waiting time for all 10 patients combined was 3030 minutes, and the median waiting time of the 3 patients logged on Tuesday was 3434 minutes. If one of the patients logged on Tuesday had a waiting time of 2222 minutes and another had a waiting time of 3434 minutes, what was the waiting time, in minutes, of the third patient logged on Tuesday?

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Cevap: 38 minutes

Cevap

38 minutes
The option stating 38 minutes is correct because the total sum for all 10 patients combined is 10×30=30010 \times 30 = 300 minutes. The sum of Monday's 7 patient waiting times is 12+41+24+19+33+28+49=20612 + 41 + 24 + 19 + 33 + 28 + 49 = 206 minutes. Therefore, Tuesday's 3 patients must sum to 300206=94300 - 206 = 94 minutes. Since two of Tuesday's patients waited 22 minutes and 34 minutes, the third patient's waiting time is 94(22+34)=3894 - (22 + 34) = 38 minutes. Ordering the three Tuesday times (22,34,38)(22, 34, 38) confirms the median is indeed 34 minutes.

Adım Adım Çözüm

1
Calculate the total sum of waiting times for all 10 patients.
Total sum = 10×30=30010 \times 30 = 300 minutes.
The arithmetic mean of nn values is Sum/n\text{Sum} / n, so Sum=n×Mean\text{Sum} = n \times \text{Mean}.
2
Calculate the sum of waiting times for the 7 patients logged on Monday.
Monday sum = 12+41+24+19+33+28+49=20612 + 41 + 24 + 19 + 33 + 28 + 49 = 206 minutes.
Adding the individual values gives the total Monday waiting time.
3
Find the total sum of waiting times for the 3 patients logged on Tuesday.
Tuesday sum = 300206=94300 - 206 = 94 minutes.
Subtracting Monday's total from the combined total yields Tuesday's total.
4
Determine the third patient's waiting time on Tuesday using the given values.
Third patient's waiting time = 94(22+34)=9456=3894 - (22 + 34) = 94 - 56 = 38 minutes.
The sum of the 3 Tuesday patients is 94 minutes, and two known patients waited 22 and 34 minutes. Note that the sorted set (22,34,38)(22, 34, 38) has a median of 34 minutes, consistent with the problem statement.

Anahtar Kavram

Combined Mean and Median Properties
Soru 1798Soru

A municipal water treatment facility uses two intake pipes, Pipe XX and Pipe YY, to fill a main reservoir. Operating alone at its constant rate, Pipe XX can fill the empty reservoir in 1010 hours. Operating alone at its constant rate, Pipe YY can fill the empty reservoir in 1515 hours. Pipe XX is turned on first and operates alone for 33 hours. Then, Pipe YY is also turned on, and both pipes operate together until the reservoir is completely full. What is the total number of hours Pipe XX operates from the moment it is turned on until the reservoir is completely filled?

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Cevap: 7.27.2

Cevap

The total number of hours Pipe X operates is 7.27.2 hours.
The correct answer is 7.27.2 hours. Pipe X completes 3/103/10 of the reservoir in 3 hours, leaving 7/107/10 of the reservoir to be filled. The combined rate of Pipe X and Pipe Y is 1/10+1/15=1/61/10 + 1/15 = 1/6 per hour. The joint time required to fill the remaining 7/107/10 is (7/10)/(1/6)=4.2(7/10) / (1/6) = 4.2 hours. Adding the initial 3 hours Pipe X worked alone yields a total of 3+4.2=7.23 + 4.2 = 7.2 hours.

Adım Adım Çözüm

1
Determine the individual hourly work rates of Pipe X and Pipe Y.
Rate of Pipe X = 110\frac{1}{10} reservoir per hour; Rate of Pipe Y = 115\frac{1}{15} reservoir per hour.
The rate is the reciprocal of the total time required to complete the job individually.
2
Calculate the fraction of the reservoir filled by Pipe X during its initial 3-hour solo operation.
Work done in first 3 hours = 3×110=3103 \times \frac{1}{10} = \frac{3}{10} of the reservoir.
Multiplying Pipe X's rate by its solo operating time gives the completed portion of the work.
3
Find the remaining fraction of the reservoir that needs to be filled.
Remaining work = 1310=7101 - \frac{3}{10} = \frac{7}{10} of the reservoir.
Subtracting the completed portion from the whole (1) leaves the uncompleted portion.
4
Determine the combined hourly rate when both pipes operate simultaneously.
Combined rate = 110+115=330+230=530=16\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} of the reservoir per hour.
Simultaneous operation rates are additive.
5
Calculate the time tt during which both pipes operate together to finish the remaining work.
t=71016=710×6=4210=4.2t = \frac{\frac{7}{10}}{\frac{1}{6}} = \frac{7}{10} \times 6 = \frac{42}{10} = 4.2 hours.
Dividing the remaining work by the combined rate yields the joint operation time.
6
Add Pipe X's solo time to the joint operation time to get Pipe X's total operating time.
Total time = 3+4.2=7.23 + 4.2 = 7.2 hours.
Pipe X was active during both the initial solo period and the joint operating period.

Anahtar Kavram

Combined Rate and Staggered Work Modeling
Tahmini Süre:2m 0s
Soru 1799Soru

A technology committee has a pool of 88 available guest speakers consisting of 55 computer scientists and 33 data privacy experts. A 44-person panel is to be formed from this pool. Which of the following statements regarding the possible panel selections must be true? Select all such statements.

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Cevap: The total number of different 44-person panels that can be formed without any restriction on specialty is equal to 7070.; The number of different 44-person panels that consist of exactly 22 computer scientists and 22 data privacy experts is equal to 3030.; The number of different 44-person panels containing at least 11 data privacy expert is equal to 6565.

Cevap

The correct statements are those asserting that total unrestricted panels equal 70, panels with exactly 2 computer scientists and 2 privacy experts equal 30, and panels with at least 1 privacy expert equal 65.
The total unrestricted 4-person panels from 8 speakers is 8C4 = 70. Selecting 2 computer scientists (5C2 = 10) and 2 privacy experts (3C2 = 3) yields 10 × 3 = 30 panels. Using complementary counting, panels with at least 1 privacy expert equal total panels (70) minus panels composed entirely of computer scientists (5C4 = 5), giving 70 - 5 = 65.

Adım Adım Çözüm

1
Calculate unrestricted combinations of 4 speakers out of 8
(84)=8×7×6×54×3×2×1=70\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70
Selection order does not matter for committee membership, so we use combinations.
2
Calculate combinations with restricted counts (2 CS and 2 PE)
(52)×(32)=10×3=30\binom{5}{2} \times \binom{3}{2} = 10 \times 3 = 30
By the Fundamental Counting Principle, independent choices for each subgroup are multiplied.
3
Calculate combinations with 'at least 1' condition using the complementary counting method
Total panels minus panels with 0 PE: 70(54)=705=6570 - \binom{5}{4} = 70 - 5 = 65
Subtracting outcomes that violate the constraint from total possible outcomes simplifies 'at least 1' calculations.
4
Evaluate the permutation and impossible constraint statements
Arranging 4 CS out of 5 gives P(5,4)=12020P(5,4) = 120 \neq 20. A 4-person panel with 0 CS requires 4 PE out of 3, which gives 010 \neq 1.
Arrangement requires permutations, and panel requirements exceeding available pool members yield zero valid groups.

Anahtar Kavram

Combinations, Permutations, and Complementary Counting Principle
Soru 1800Soru

A quality control analyst evaluates a manufacturing process in which two specific types of flaws, Flaw XX and Flaw YY, can occur on produced glass panels. The probability that a randomly selected panel has Flaw XX is P(X)=0.20P(X) = 0.20, and the probability that it has Flaw YY is P(Y)=0.30P(Y) = 0.30. The analyst confirms that Flaw XX and Flaw YY are mutually exclusive events.

Which of the following statements MUST be true regarding these two flaw types? Select all such statements.

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Cevap: The probability that a randomly selected panel has at least one of the two flaws is 0.500.50.; The conditional probability of a panel having Flaw XX given that it has Flaw YY, P(XY)P(X \mid Y), is equal to 00.

Cevap

The statement that the probability of having at least one flaw is 0.50 and the statement that the conditional probability P(X | Y) is 0 are both true.
Because Flaw XX and Flaw YY are mutually exclusive, their intersection P(XY)=0P(X \cap Y) = 0. By the addition rule, the probability of at least one flaw is P(XY)=P(X)+P(Y)=0.20+0.30=0.50P(X \cup Y) = P(X) + P(Y) = 0.20 + 0.30 = 0.50. Furthermore, the conditional probability P(XY)=P(XY)P(Y)=00.30=0P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0}{0.30} = 0. Thus, both the statement claiming the union probability is 0.500.50 and the statement claiming the conditional probability is 00 are correct.

Adım Adım Çözüm

1
Analyze the definition of mutually exclusive events
Since Flaw XX and Flaw YY are mutually exclusive, they cannot occur simultaneously on the same panel. Therefore, P(XY)=0P(X \cap Y) = 0.
By definition, mutually exclusive events have an intersection probability of zero.
2
Calculate the union probability P(X or Y)
Using the addition rule for mutually exclusive events: P(XY)=P(X)+P(Y)P(XY)=0.20+0.300=0.50P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = 0.20 + 0.30 - 0 = 0.50.
The probability of at least one event occurring is the sum of their individual probabilities when the intersection is zero.
3
Calculate the conditional probability P(X | Y)
P(XY)=P(XY)P(Y)=00.30=0P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0}{0.30} = 0.
If Flaw YY is known to occur, Flaw XX cannot occur due to mutual exclusivity.
4
Evaluate independence between the two events
For independence, P(XY)P(X \cap Y) must equal P(X)×P(Y)=0.20×0.30=0.06P(X) \times P(Y) = 0.20 \times 0.30 = 0.06. Since 00.060 \neq 0.06, the events are dependent.
Two events with non-zero probabilities that are mutually exclusive are always dependent because the occurrence of one guarantees the non-occurrence of the other.

Anahtar Kavram

Mutually Exclusive vs. Independent Events
ÖncekiSayfa 90 / 107Sonraki
Tüm alıştırma soruları — GRE General Test | Examkin