Tüm alıştırma soruları

1526 soru

Soru 141Soru

Evaluate the limit limx2x24x2\lim_{x \to 2} \frac{x^2 - 4}{x - 2}. What is the numerical value of this limit?

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Cevap: 4

Cevap

The value of the limit is 44.
Direct substitution of x=2x = 2 produces the indeterminate form 00\frac{0}{0}. Factoring the numerator gives (x2)(x+2)x2\frac{(x-2)(x+2)}{x-2}. Canceling the non-zero factor (x2)(x-2) simplifies the expression to x+2x+2. Evaluating the limit as xx approaches 22 yields 2+2=42 + 2 = 4.

Adım Adım Çözüm

1
Check the form by direct substitution of x=2x = 2
Obtained the indeterminate form 00\frac{0}{0}
Direct substitution results in division by zero, requiring algebraic simplification.
2
Factor the polynomial in the numerator
x24=(x2)(x+2)x^2 - 4 = (x - 2)(x + 2)
Difference of two squares factorization allows canceling common terms.
3
Cancel the common factor (x2)(x - 2)
(x2)(x+2)x2=x+2\frac{(x - 2)(x + 2)}{x - 2} = x + 2
For x2x \neq 2, division by (x2)(x - 2) is valid.
4
Evaluate the simplified limit as x2x \to 2
2+2=42 + 2 = 4
Substitute x=2x = 2 directly into the continuous polynomial x+2x + 2.

Anahtar Kavram

Evaluating indeterminate limits of the form 00\frac{0}{0} via factorization
Soru 142Soru

Three business partners share a total profit of ₦60,000 in the ratio 1:2:31 : 2 : 3. What is the share of the partner who receives the largest portion, in naira?

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Cevap: 30000

Cevap

The share of the partner receiving the largest portion is ₦30,000.
Sum the parts of the ratio (1+2+3=61 + 2 + 3 = 6). The largest share corresponds to 3 parts out of 6. Calculating 36×60,000\frac{3}{6} \times 60,000 yields 30,000 naira.

Adım Adım Çözüm

1
Calculate the sum of all ratio parts
1 + 2 + 3 = 6 parts
The total amount is divided into equal parts represented by the sum of the ratio numbers.
2
Determine the monetary value of a single part
₦60,000 / 6 = ₦10,000 per part
Dividing the total sum by the total number of parts gives the value of one unit part.
3
Calculate the largest share corresponding to 3 parts
3 × ₦10,000 = ₦30,000
The largest portion of the ratio is 3 parts.

Anahtar Kavram

Direct Ratio Sharing
Soru 143Soru

Two independent events, AA and BB, have probabilities P(A)=0.4P(A) = 0.4 and P(B)=0.5P(B) = 0.5. What is the probability that both event AA and event BB occur, P(AB)P(A \cap B)?

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Cevap: 0.2

Cevap

The probability that both events occur is 0.2.
For independent events, the joint probability of both events occurring simultaneously is found by multiplying their individual probabilities: P(AB)=P(A)×P(B)=0.4×0.5=0.2P(A \cap B) = P(A) \times P(B) = 0.4 \times 0.5 = 0.2.

Adım Adım Çözüm

1
Identify event independence and the required probability operation
Events AA and BB are independent, and the question requires calculating their intersection P(AB)P(A \cap B).
The problem explicitly states that the events are independent.
2
Apply the multiplication law for independent events
P(AB)=P(A)×P(B)P(A \cap B) = P(A) \times P(B)
For any two independent events, the probability of both occurring together is the product of their individual probabilities.
3
Perform the multiplication
P(AB)=0.4×0.5=0.2P(A \cap B) = 0.4 \times 0.5 = 0.2
Multiplying 0.4 by 0.5 gives 0.2.

Anahtar Kavram

Multiplication Law of Probability for Independent Events
Soru 144Soru

The sum of the first nn terms of an arithmetic progression is given by Sn=2n2+3nS_n = 2n^2 + 3n. What is the value of the 7th7^{\text{th}} term of the progression?

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Cevap: 29

Cevap

The 7th7^{\text{th}} term of the arithmetic progression is 2929.
For any sequence, the nn-th term is found using Tn=SnSn1T_n = S_n - S_{n-1}. Substituting n=7n = 7 gives S7=2(7)2+3(7)=119S_7 = 2(7)^2 + 3(7) = 119 and S6=2(6)2+3(6)=90S_6 = 2(6)^2 + 3(6) = 90. Thus, T7=11990=29T_7 = 119 - 90 = 29.

Adım Adım Çözüm

1
State the relationship between the nn-th term TnT_n and the sum of first nn terms SnS_n
Tn=SnSn1T_n = S_n - S_{n-1}
The sum of the first nn terms minus the sum of the first n1n-1 terms equals the nn-th term.
2
Calculate the sum of the first 7 terms (S7S_7)
S7=2(7)2+3(7)=119S_7 = 2(7)^2 + 3(7) = 119
Substitute n=7n = 7 into the sum formula Sn=2n2+3nS_n = 2n^2 + 3n.
3
Calculate the sum of the first 6 terms (S6S_6)
S6=2(6)2+3(6)=90S_6 = 2(6)^2 + 3(6) = 90
Substitute n=6n = 6 into the sum formula Sn=2n2+3nS_n = 2n^2 + 3n.
4
Compute the 7th7^{\text{th}} term (T7T_7)
T7=11990=29T_7 = 119 - 90 = 29
Subtract S6S_6 from S7S_7.

Anahtar Kavram

Relationship between the nth term and the sum of first n terms of an AP
Soru 145Soru

The mean mark of a student in 66 tests is 1414. When the highest and lowest marks, which differ by 1212, are excluded, the mean mark of the remaining 44 tests becomes 13.513.5. What is the highest mark?

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Cevap: 21

Cevap

The highest mark is 21.
The total sum of all 6 tests is 6×14=846 \times 14 = 84. When the highest (HH) and lowest (LL) marks are removed, the total sum of the remaining 4 tests is 4×13.5=544 \times 13.5 = 54. The sum of the excluded marks is H+L=8454=30H + L = 84 - 54 = 30. Knowing that their difference is HL=12H - L = 12, we add the two equations to get 2H=422H = 42, which gives H=21H = 21.

Adım Adım Çözüm

1
Calculate the total sum of all 6 test marks
Sum of 6 marks = 6×14=846 \times 14 = 84
The sum of data values equals the mean multiplied by the number of items.
2
Calculate the sum of the remaining 4 test marks
Sum of 4 marks = 4×13.5=544 \times 13.5 = 54
Multiplying the new mean by 4 gives the sum of the test scores excluding the highest and lowest values.
3
Determine the combined sum of the highest (H) and lowest (L) marks
H + L = 84 - 54 = 30
Subtracting the sum of the 4 remaining marks from the total initial sum yields the sum of the two excluded marks.
4
Solve for H using the simultaneous linear equations
H = 21
Adding H + L = 30 and H - L = 12 gives 2H = 42, which solves to H = 21.

Anahtar Kavram

Mean of Ungrouped Data and Handling Excluded Values
Soru 146Soru

A convex polygon has 5454 diagonals. How many distinct triangles can be formed by joining any three of its vertices?

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Cevap: 220

Cevap

220 distinct triangles
Solving the equation for the number of diagonals n(n3)2=54\frac{n(n-3)}{2} = 54 yields n=12n = 12 vertices. The number of triangles that can be formed by selecting any 3 of these 12 vertices is given by (123)=12×11×106=220\binom{12}{3} = \frac{12 \times 11 \times 10}{6} = 220.

Adım Adım Çözüm

1
Determine the number of vertices nn of the polygon using the diagonals formula.
n=12n = 12
The number of diagonals DD in an nn-sided convex polygon is given by D=(n2)n=n(n3)2D = \binom{n}{2} - n = \frac{n(n-3)}{2}. Setting n(n3)2=54\frac{n(n-3)}{2} = 54 gives n23n108=0n^2 - 3n - 108 = 0. Factoring (n12)(n+9)=0(n - 12)(n + 9) = 0 yields n=12n = 12 since the number of vertices must be positive.
2
Calculate the number of distinct triangles formed by choosing 3 vertices from 12.
220220
Each set of 3 distinct vertices forms one unique triangle. The order in which the vertices are chosen does not matter, so we use combinations: (123)=12×11×103×2×1=220\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220.

Anahtar Kavram

Combinations applied to geometric figures and polygon properties
Tahmini Süre:2m 0s
Soru 147Soru

Find the smallest positive integer nn that simultaneously satisfies the linear modular congruences 3n5(mod13)3n \equiv 5 \pmod{13} and 4n2(mod9)4n \equiv 2 \pmod{9}.

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Cevap: 32

Cevap

The smallest positive integer satisfying both congruences is 32.
Solving the first congruence 3n5(mod13)3n \equiv 5 \pmod{13} gives n6(mod13)n \equiv 6 \pmod{13} (since 3×91(mod13)3 \times 9 \equiv 1 \pmod{13} and 5×9=456(mod13)5 \times 9 = 45 \equiv 6 \pmod{13}), which means nn can be written as 13k+613k + 6. Substituting this into the second congruence 4n2(mod9)4n \equiv 2 \pmod{9} yields 4(13k+6)2(mod9)    52k+242(mod9)4(13k + 6) \equiv 2 \pmod{9} \implies 52k + 24 \equiv 2 \pmod{9}. Reducing the coefficients modulo 9 gives 7k+62(mod9)    7k45(mod9)7k + 6 \equiv 2 \pmod{9} \implies 7k \equiv -4 \equiv 5 \pmod{9}. Multiplying by 4 (the modular inverse of 7 modulo 9) yields k202(mod9)k \equiv 20 \equiv 2 \pmod{9}. Setting k=2k = 2 yields the smallest positive integer n=13(2)+6=32n = 13(2) + 6 = 32.

Adım Adım Çözüm

1
Solve the first modular congruence 3n5(mod13)3n \equiv 5 \pmod{13} for nn
n6(mod13)n \equiv 6 \pmod{13}, which implies n=13k+6n = 13k + 6
Multiplying both sides by the modular inverse of 3 modulo 13 (which is 9) isolates nn.
2
Solve the second modular congruence 4n2(mod9)4n \equiv 2 \pmod{9} for nn
n5(mod9)n \equiv 5 \pmod{9}
Multiplying both sides by the modular inverse of 4 modulo 9 (which is 7) isolates nn.
3
Substitute n=13k+6n = 13k + 6 into n5(mod9)n \equiv 5 \pmod{9} and simplify modulo 9
4k8(mod9)4k \equiv 8 \pmod{9}
Reducing 13 modulo 9 yields 4k4k, and subtracting 6 from 5 yields 18(mod9)-1 \equiv 8 \pmod{9}.
4
Solve for kk and calculate the smallest positive integer nn
k2(mod9)k \equiv 2 \pmod{9}, giving n=13(2)+6=32n = 13(2) + 6 = 32
Setting the integer parameter kk to its minimum non-negative value 22 provides the smallest positive integer solution.

Anahtar Kavram

System of Linear Modular Congruences and Modular Inverses
Soru 148Soru

What is the 10th10^{\text{th}} term of the arithmetic progression 3,7,11,15,3, 7, 11, 15, \dots?

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Cevap: 39

Cevap

The 10th10^{\text{th}} term of the arithmetic progression is 3939.
By applying the nthn^{\text{th}} term formula for an arithmetic progression Tn=a+(n1)dT_n = a + (n - 1)d with first term a=3a = 3, common difference d=4d = 4, and term index n=10n = 10, the calculation yields T10=3+(101)×4=3+36=39T_{10} = 3 + (10 - 1) \times 4 = 3 + 36 = 39.

Adım Adım Çözüm

1
Identify the key parameters of the arithmetic progression from the given sequence.
First term a=3a = 3, common difference d=73=4d = 7 - 3 = 4, and number of terms n=10n = 10.
These parameters are required to use the nthn^{\text{th}} term formula of an A.P.
2
Substitute the values into the formula Tn=a+(n1)dT_n = a + (n - 1)d.
T10=3+(101)×4T_{10} = 3 + (10 - 1) \times 4
The formula relates the nthn^{\text{th}} term to the first term, common difference, and term position.
3
Evaluate the mathematical expression.
T10=3+9×4=3+36=39T_{10} = 3 + 9 \times 4 = 3 + 36 = 39
Perform multiplication before addition according to standard order of operations.

Anahtar Kavram

nth term of an Arithmetic Progression
Soru 149Soru

What is the measure, in degrees, of each interior angle of a regular octagon (an 8-sided regular polygon)?

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Cevap: 135

Cevap

Each interior angle of a regular octagon measures 135 degrees.
The sum of the interior angles of a polygon with nn sides is (n2)×180(n - 2) \times 180^\circ. For a regular octagon (n=8n = 8), the total interior angle sum is (82)×180=1080(8 - 2) \times 180^\circ = 1080^\circ. Since all 8 interior angles of a regular octagon are congruent, dividing the total sum by 8 yields 135135^\circ for each interior angle.

Adım Adım Çözüm

1
Find the sum of all interior angles of the regular octagon.
Sum of interior angles = (82)×180=6×180=1080(8 - 2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ.
The sum of interior angles for any nn-sided polygon is (n2)×180(n - 2) \times 180^\circ.
2
Calculate the measure of a single interior angle.
Interior angle = 10808=135\frac{1080^\circ}{8} = 135^\circ.
In a regular polygon, all interior angles are equal in measure.

Anahtar Kavram

Interior Angle of a Regular Polygon
Soru 150Soru

Given the simultaneous equations x+2y=7x + 2y = 7 and x2+3xy+y2=19x^2 + 3xy + y^2 = 19, find the product of all possible values of yy that satisfy the system.

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Cevap: -30

Cevap

The product of all possible values of y that satisfy the system is -30.
Rearranging the linear equation gives x = 7 - 2y. Substituting this into the quadratic equation x^2 + 3xy + y^2 = 19 produces (7 - 2y)^2 + 3(7 - 2y)y + y^2 = 19. Expanding and combining like terms yields y^2 + 7y - 30 = 0. Solving for y gives y = 3 and y = -10. Multiplying these values together gives a product of -30.

Adım Adım Çözüm

1
Isolate x in the linear equation
x = 7 - 2y
Expressing one variable in terms of the other enables substitution into the quadratic equation.
2
Substitute x into the quadratic equation and expand
(7 - 2y)^2 + 3(7 - 2y)y + y^2 = 19
This reduces the system to a single quadratic equation in terms of y.
3
Simplify the resulting expression into standard quadratic form
y^2 + 7y - 30 = 0
Expanding yields (49 - 28y + 4y^2) + (21y - 6y^2) + y^2 = 19, which reduces to y^2 + 7y - 30 = 0.
4
Calculate the product of the roots of y
y_1 * y_2 = -30
By Vieta's formulas, the product of roots for y^2 + ay + b = 0 is b/1 = -30 (or factoring gives y = 3 and y = -10, with product 3 * (-10) = -30).

Anahtar Kavram

Solving simultaneous linear and quadratic equations via substitution and applying quadratic root properties
Tahmini Süre:2m 0s
Soru 151Soru

If y=(2x+1)3y = (2x + 1)^3, find the value of dydx\frac{dy}{dx} at x=1x = 1.

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Cevap: 54

Cevap

54
Applying the chain rule gives dydx=3(2x+1)22=6(2x+1)2\frac{dy}{dx} = 3(2x + 1)^2 \cdot 2 = 6(2x + 1)^2. Evaluating at x=1x = 1 gives 6(3)2=546(3)^2 = 54.

Adım Adım Çözüm

1
Differentiate y=(2x+1)3y = (2x + 1)^3 using the chain rule.
dydx=6(2x+1)2\frac{dy}{dx} = 6(2x + 1)^2
According to the chain rule, ddx[un]=nun1dudx\frac{d}{dx}[u^n] = n u^{n-1} \cdot \frac{du}{dx}, where u=2x+1u = 2x + 1 and dudx=2\frac{du}{dx} = 2.
2
Evaluate the derivative at x=1x = 1.
dydxx=1=54\frac{dy}{dx}\Big|_{x=1} = 54
Substituting x=1x = 1 into 6(2x+1)26(2x + 1)^2 yields 6(3)2=546(3)^2 = 54.

Anahtar Kavram

Chain Rule for Differentiation
Soru 152Soru
Find the real value of xx that satisfies the exponential equation 52x1×25x+1=125x+25^{2x - 1} \times 25^{x + 1} = 125^{x + 2}
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Cevap: 5

Cevap

The value of xx is 5.
By converting all terms to base 5 (25=5225 = 5^2 and 125=53125 = 5^3), the equation becomes 52x1×52(x+1)=53(x+2)5^{2x-1} \times 5^{2(x+1)} = 5^{3(x+2)}. Simplifying exponents gives 52x1×52x+2=53x+65^{2x-1} \times 5^{2x+2} = 5^{3x+6}. Adding the left-hand powers results in 54x+1=53x+65^{4x+1} = 5^{3x+6}. Equating the exponents yields 4x+1=3x+64x + 1 = 3x + 6, leading directly to x=5x = 5.

Adım Adım Çözüm

1
Express all terms with a common base of 5
52x1×52x+2=53x+65^{2x-1} \times 5^{2x+2} = 5^{3x+6}
Since 25=5225 = 5^2 and 125=53125 = 5^3, using index laws (am)n=amn(a^m)^n = a^{mn} allows all expressions to share base 5.
2
Apply the product rule of indices on the left side
54x+1=53x+65^{4x+1} = 5^{3x+6}
According to the product law am×an=am+na^m \times a^n = a^{m+n}, the powers are added: (2x1)+(2x+2)=4x+1(2x-1) + (2x+2) = 4x+1.
3
Equate powers of equal bases to solve for x
x=5x = 5
Since bases are equal, exponents must be equal: 4x+1=3x+6    4x3x=61    x=54x + 1 = 3x + 6 \implies 4x - 3x = 6 - 1 \implies x = 5.

Anahtar Kavram

Indices and Laws of Indices
Soru 153Soru

In a probability experiment, two fair six-sided dice were rolled 180180 times, yielding an experimental probability of 518\frac{5}{18} for obtaining a sum divisible by 33. If mm additional consecutive rolls were conducted and every single one resulted in a sum divisible by 33, the updated overall experimental probability equaled the theoretical probability that the absolute difference between the numbers shown on two fair six-sided dice is at most 11. Calculate the value of mm.

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Cevap: 54

Cevap

54
The initial number of successful trials is 180×518=50180 \times \frac{5}{18} = 50. The theoretical probability of rolling two dice with an absolute difference of at most 11 is calculated by counting 66 outcomes with difference 00 and 1010 outcomes with difference 11, giving 1636=49\frac{16}{36} = \frac{4}{9}. Equating the updated experimental probability 50+m180+m\frac{50 + m}{180 + m} to 49\frac{4}{9} yields 9(50+m)=4(180+m)9(50 + m) = 4(180 + m), which simplifies to 5m=2705m = 270 or m=54m = 54.

Adım Adım Çözüm

1
Calculate the initial number of successful trials from the given experimental probability.
Initial successful outcomes = 180×518=50180 \times \frac{5}{18} = 50.
Experimental probability is defined as the number of successful trials divided by the total number of trials.
2
Calculate the theoretical probability that the absolute difference between two rolled six-sided dice is at most 1.
Favorable outcomes = 16, so P(theoretical)=1636=49P(\text{theoretical}) = \frac{16}{36} = \frac{4}{9}.
Outcomes with difference 0: (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)(1,1), (2,2), (3,3), (4,4), (5,5), (6,6) (6 outcomes). Outcomes with difference 1: (1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5)(1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6), (6,5) (10 outcomes). Total sample space =6×6=36= 6 \times 6 = 36.
3
Formulate the algebraic equation relating the updated experimental probability to the theoretical probability.
50+m180+m=49\frac{50 + m}{180 + m} = \frac{4}{9}.
Adding mm consecutive successful rolls increases both the number of successful outcomes (to 50+m50 + m) and the total number of trials (to 180+m180 + m).
4
Solve the equation for mm.
9(50+m)=4(180+m)    450+9m=720+4m    5m=270    m=549(50 + m) = 4(180 + m) \implies 450 + 9m = 720 + 4m \implies 5m = 270 \implies m = 54.
Cross-multiplication converts the rational expression into a linear equation.

Anahtar Kavram

Experimental and Theoretical Probability Synthesis
Tahmini Süre:3m 0s
Soru 154Soru

Pump A can fill a water reservoir in 66 hours, while a drain pipe can empty the full reservoir in 1515 hours. Pump A is switched on to fill an empty reservoir while the drain pipe is accidentally left open. After 33 hours, an identical pump, Pump B, is also switched on to assist Pump A while the drain pipe remains open. How many total hours will it take for the reservoir to become completely full?

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Cevap: 5.625

Cevap

The total time required to fill the reservoir completely is 5.6255.625 hours.
To solve multi-stage work and rate problems involving opposing forces (filling vs. draining), calculate the net rate of change per unit of time for each stage. In stage one, Pump A adds 16\frac{1}{6} while the drain removes 115\frac{1}{15}, giving a net rate of 110\frac{1}{10} per hour. In 3 hours, 310\frac{3}{10} of the reservoir is filled, leaving 710\frac{7}{10}. In stage two, adding identical Pump B increases the filling rate to 2×16=132 \times \frac{1}{6} = \frac{1}{3}. Subtracting the drain rate 115\frac{1}{15} yields a net rate of 415\frac{4}{15} per hour. Dividing the remaining 710\frac{7}{10} by 415\frac{4}{15} gives 2.6252.625 hours. Adding the initial 3 hours yields a total of 5.6255.625 hours.

Adım Adım Çözüm

1
Determine individual hourly rates
Pump A rate = +16+\frac{1}{6} reservoir/hr, Drain rate = 115-\frac{1}{15} reservoir/hr
Rate is the reciprocal of the time required to complete the full job.
2
Calculate net rate and progress for the first 3 hours
Net rate = 16115=110\frac{1}{6} - \frac{1}{15} = \frac{1}{10} reservoir/hr. Progress in 3 hours = 3×110=3103 \times \frac{1}{10} = \frac{3}{10} of the reservoir.
Only Pump A and the drain pipe are active during the initial 3-hour period.
3
Calculate remaining fraction of reservoir to be filled
Remaining portion = 1310=7101 - \frac{3}{10} = \frac{7}{10}
The total capacity of the reservoir is represented by 11 whole unit.
4
Calculate the combined rate after Pump B is added
New net rate = 16+16115=13115=415\frac{1}{6} + \frac{1}{6} - \frac{1}{15} = \frac{1}{3} - \frac{1}{15} = \frac{4}{15} reservoir/hr
Pump B is identical to Pump A, so its rate is also 16\frac{1}{6} reservoir/hr.
5
Find additional time needed and total time elapsed
Additional time = 7/104/15=710×154=218=2.625\frac{7/10}{4/15} = \frac{7}{10} \times \frac{15}{4} = \frac{21}{8} = 2.625 hours. Total time = 3+2.625=5.6253 + 2.625 = 5.625 hours.
Time equals remaining work divided by combined net rate, then added to elapsed time.

Anahtar Kavram

Work-Rate and Simultaneous Operations (Combined Filling and Emptying Rates)
Tahmini Süre:2m 30s
Soru 155Soru

The sum of the first four terms of an arithmetic progression (A.P.) is 3232, and the sum of the next four terms is 9696. What is the common difference of the progression?

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Cevap: 4

Cevap

The common difference of the arithmetic progression is 44.
The sum of the first four terms yields 2a+3d=162a + 3d = 16. The sum of the first eight terms is 32+96=12832 + 96 = 128, which gives 2a+7d=322a + 7d = 32. Subtracting these two linear equations gives 4d=164d = 16, leading to d=4d = 4.

Adım Adım Çözüm

1
Formulate an equation for the sum of the first 4 terms.
2a+3d=162a + 3d = 16
The sum of the first nn terms of an A.P. is Sn=n2(2a+(n1)d)S_n = \frac{n}{2}(2a + (n-1)d). Substituting n=4n = 4 and S4=32S_4 = 32 gives 2(2a+3d)=322(2a + 3d) = 32, which simplifies to 2a+3d=162a + 3d = 16.
2
Formulate an equation for the sum of the first 8 terms.
2a+7d=322a + 7d = 32
The total sum of the first 8 terms is the sum of the first 4 terms plus the sum of the next 4 terms (S8=32+96=128S_8 = 32 + 96 = 128). Substituting n=8n = 8 gives 4(2a+7d)=1284(2a + 7d) = 128, which simplifies to 2a+7d=322a + 7d = 32.
3
Solve the system of simultaneous linear equations for dd.
d=4d = 4
Subtracting (2a+3d=16)(2a + 3d = 16) from (2a+7d=32)(2a + 7d = 32) eliminates 2a2a, resulting in 4d=164d = 16, which yields d=4d = 4.

Anahtar Kavram

Sum of an Arithmetic Progression
Soru 156Soru

In an agricultural experiment, two crop varieties, XX and YY, are tested independently for germination under drought conditions. The probability that variety XX germinates is 0.650.65, and the probability that at least one of the two varieties germinates is 0.860.86. What is the probability that variety YY germinates?

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Cevap: 0.6

Cevap

The probability that variety Y germinates is 0.6
Using the law of addition for independent events P(XY)=P(X)+P(Y)P(X)P(Y)P(X \cup Y) = P(X) + P(Y) - P(X)P(Y), substituting P(X)=0.65P(X) = 0.65 and P(XY)=0.86P(X \cup Y) = 0.86 gives 0.86=0.65+0.35P(Y)0.86 = 0.65 + 0.35 P(Y), which yields P(Y)=0.6P(Y) = 0.6.

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1
Apply the general addition law for two probability events
P(XY)=P(X)+P(Y)P(XY)P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)
The addition law relates the union, individual probabilities, and intersection of compound events.
2
Express the intersection using the law of multiplication for independent events
P(XY)=P(X)×P(Y)=0.65×P(Y)P(X \cap Y) = P(X) \times P(Y) = 0.65 \times P(Y)
Because germination of variety X and variety Y are independent events.
3
Substitute given values into the combined probability formula and solve for P(Y)P(Y)
0.86=0.65+P(Y)0.65P(Y)    0.21=0.35P(Y)    P(Y)=0.60.86 = 0.65 + P(Y) - 0.65 P(Y) \implies 0.21 = 0.35 P(Y) \implies P(Y) = 0.6
Isolating the unknown probability P(Y)P(Y) yields the correct decimal value.

Anahtar Kavram

Compound Probability Laws and Independent Events
Soru 157Soru

The table below shows the speed distribution (in km/h) recorded for 5050 vehicles passing a police checkpoint on a highway:

Speed Interval (km/h)Frequency (ff)
404940 - 4966
505950 - 591212
606960 - 691818
707970 - 791010
808980 - 8944

Calculate the mean speed of the vehicles.

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Cevap: 63.3

Cevap

The mean speed of the vehicles is 63.3 km/h63.3\text{ km/h}.
The mean of a grouped frequency distribution is computed using the formula xˉ=fxf\bar{x} = \frac{\sum fx}{\sum f}. The midpoints (xx) for the intervals are 44.5,54.5,64.5,74.5,44.5, 54.5, 64.5, 74.5, and 84.584.5. Multiplying these midpoints by their respective frequencies yields products totaling 31653165. Dividing this total by the total number of vehicles (5050) gives a mean speed of 63.3 km/h63.3\text{ km/h}.

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1
Find the class midpoints (xx) for each speed interval
Midpoints are x1=44.5x_1 = 44.5, x2=54.5x_2 = 54.5, x3=64.5x_3 = 64.5, x4=74.5x_4 = 74.5, and x5=84.5x_5 = 84.5.
The midpoint of a grouped class interval is calculated as Lower limit+Upper limit2\frac{\text{Lower limit} + \text{Upper limit}}{2}.
2
Calculate the product fxfx for each class
6×44.5=2676 \times 44.5 = 267, 12×54.5=65412 \times 54.5 = 654, 18×64.5=116118 \times 64.5 = 1161, 10×74.5=74510 \times 74.5 = 745, and 4×84.5=3384 \times 84.5 = 338.
Multiplying class midpoint by frequency estimates the total contribution of all items within that interval.
3
Find total frequency f\sum f and sum of products fx\sum fx
f=6+12+18+10+4=50\sum f = 6 + 12 + 18 + 10 + 4 = 50 and fx=267+654+1161+745+338=3165\sum fx = 267 + 654 + 1161 + 745 + 338 = 3165.
Summing frequencies gives total sample size, and summing fxfx gives estimated grand total.
4
Compute the estimated mean speed
xˉ=fxf=316550=63.3 km/h\bar{x} = \frac{\sum fx}{\sum f} = \frac{3165}{50} = 63.3\text{ km/h}.
The standard formula for the mean of grouped frequency data is xˉ=fxf\bar{x} = \frac{\sum fx}{\sum f}.

Anahtar Kavram

Grouped Mean Calculation
Soru 158Soru

What is the determinant of the 2×22 \times 2 matrix P=(5234)P = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}?

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Cevap: 14

Cevap

The determinant of matrix PP is 1414.
For any 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, the determinant is given by adbcad - bc. Substituting a=5a=5, b=2b=2, c=3c=3, and d=4d=4 yields (5×4)(2×3)=206=14(5 \times 4) - (2 \times 3) = 20 - 6 = 14.

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1
Apply the determinant formula for a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}, which is det(P)=adbc\det(P) = ad - bc.
\det(P) = (5)(4) - (2)(3)
The determinant of a 2×22 \times 2 matrix is defined as the product of the main diagonal elements minus the product of the off-diagonal elements.
2
Evaluate the arithmetic expression.
20 - 6 = 14
Perform multiplication followed by subtraction to get the final determinant value.

Anahtar Kavram

Determinant of a 2x2 Matrix
Tahmini Süre:45s
Soru 159Soru

If y=5sinx+e3xy = 5\sin x + e^{3x}, what is the value of dydx\frac{dy}{dx} at x=0x = 0?

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Cevap: 8

Cevap

The value of the derivative at x=0x = 0 is 8.
Differentiating y=5sinx+e3xy = 5\sin x + e^{3x} yields dydx=5cosx+3e3x\frac{dy}{dx} = 5\cos x + 3e^{3x}. Substituting x=0x = 0 gives 5cos(0)+3e0=5(1)+3(1)=85\cos(0) + 3e^{0} = 5(1) + 3(1) = 8.

Adım Adım Çözüm

1
Differentiate each term of the function y=5sinx+e3xy = 5\sin x + e^{3x} with respect to xx.
\frac{dy}{dx} = 5\cos x + 3e^{3x}
The derivative of sinx\sin x is cosx\cos x, and applying the chain rule to e3xe^{3x} gives 3e3x3e^{3x}.
2
Evaluate the derivative dydx\frac{dy}{dx} at x=0x = 0.
5\cos(0) + 3e^0 = 5(1) + 3(1) = 8
Evaluating trigonometric and exponential functions at zero gives cos(0)=1\cos(0) = 1 and e0=1e^0 = 1.

Anahtar Kavram

Differentiation of trigonometric and exponential functions and evaluation of derivatives at specific points.
Soru 160Soru

A student measures the length of a metal rod as 4.0 cm4.0\text{ cm}. If the actual length of the rod is 5.0 cm5.0\text{ cm}, what is the percentage error in the measurement?

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Cevap: 20

Cevap

The percentage error in the measurement is 20%.
The absolute error is the difference between the true length (5.0 cm5.0\text{ cm}) and the measured length (4.0 cm4.0\text{ cm}), which is 1.0 cm1.0\text{ cm}. Dividing 1.0 cm1.0\text{ cm} by the true length 5.0 cm5.0\text{ cm} gives a fractional error of 0.200.20. Expressed as a percentage, 0.20×100%=20%0.20 \times 100\% = 20\%.

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1
Find the error (difference between measured value and actual value)
Error = 5.0 cm4.0 cm=1.0 cm|5.0\text{ cm} - 4.0\text{ cm}| = 1.0\text{ cm}
Absolute error is defined as the magnitude of the difference between the actual value and the measured value.
2
Calculate the percentage error
Percentage error = 1.0 cm5.0 cm×100%=20%\frac{1.0\text{ cm}}{5.0\text{ cm}} \times 100\% = 20\%
Percentage error is the ratio of absolute error to the actual value, expressed as a percentage.

Anahtar Kavram

Percentage Error Calculation
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