Tüm alıştırma soruları

13931 soru

Soru 7521Soru

What is the result of the subtraction 52482678524_8 - 267_8 expressed in base 8?

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Cevap: 2358235_8

Cevap

The correct answer is 2358235_8.
Subtracting 2678267_8 from 5248524_8 using direct base-8 subtraction requires borrowing 8 when a digit is smaller than the digit subtracted from it. In the units column, borrowing 1 (which equals 8) gives (4+8)7=5(4 + 8) - 7 = 5. In the eights column, the borrowed-from digit 2 becomes 1; borrowing 1 (which equals 8) gives (1+8)6=3(1 + 8) - 6 = 3. In the sixty-fours column, the borrowed-from digit 5 becomes 4; 42=24 - 2 = 2. Thus, the difference is 2358235_8.

Adım Adım Çözüm

1
Convert both octal numbers to base 10.
5248=(5×82)+(2×81)+(4×80)=320+16+4=34010524_8 = (5 \times 8^2) + (2 \times 8^1) + (4 \times 8^0) = 320 + 16 + 4 = 340_{10}, and 2678=(2×82)+(6×81)+(7×80)=128+48+7=18310267_8 = (2 \times 8^2) + (6 \times 8^1) + (7 \times 8^0) = 128 + 48 + 7 = 183_{10}.
Converting to decimal simplifies standard arithmetic operations.
2
Subtract the decimal values.
3401018310=15710340_{10} - 183_{10} = 157_{10}.
Obtain the difference in base 10.
3
Convert the decimal difference 15710157_{10} back to base 8.
157÷8=19157 \div 8 = 19 remainder 55; 19÷8=219 \div 8 = 2 remainder 33; 2÷8=02 \div 8 = 0 remainder 22. Reading remainders from bottom to top gives 2358235_8.
Successive division by 8 yields the octal representation.

Anahtar Kavram

Non-decimal base arithmetic and borrowing rules
Soru 7522Soru

The 5th5^{\text{th}} term of an arithmetic progression (AP) is 1717 and its common difference is 33. What is the first term of the progression?

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Cevap: 55

Cevap

The first term of the progression is 55.
The nthn^{\text{th}} term formula of an arithmetic progression is Tn=a+(n1)dT_n = a + (n - 1)d. Substituting T5=17T_5 = 17, n=5n = 5, and d=3d = 3 gives 17=a+4(3)=a+1217 = a + 4(3) = a + 12. Isolating aa yields a=1712=5a = 17 - 12 = 5.

Adım Adım Çözüm

1
Identify the given values and formula for the nth term of an AP.
The formula is Tn=a+(n1)dT_n = a + (n - 1)d, with T5=17T_5 = 17, n=5n = 5, and d=3d = 3.
This formula connects the nth term, the first term, the number of terms, and the common difference.
2
Substitute the known values into the equation.
17=a+(51)×3    17=a+1217 = a + (5 - 1) \times 3 \implies 17 = a + 12.
Subtracting 1 from the term index 5 gives 4, and multiplying by 3 gives 12.
3
Solve for the first term aa.
a=1712=5a = 17 - 12 = 5.
Subtracting 12 from both sides isolates aa.

Anahtar Kavram

n-th term of an Arithmetic Progression
Tahmini Süre:1m 0s
Soru 7523Soru

A straight line L1L_1 passes through the points A(2,1)A(-2, 1) and B(4,5)B(4, 5). A second line L2L_2 is perpendicular to L1L_1 and passes through the midpoint of the line segment ABAB. If L2L_2 intersects the xx-axis at point RR, and RR divides the line segment joining P(1,2)P(1, -2) and Q(k,4)Q(k, 4) internally in the ratio 1:21 : 2, what is the value of kk?

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Cevap: 77

Cevap

The value of kk is 77.
The midpoint of ABAB is (1,3)(1, 3) and the gradient of L1L_1 is 23\frac{2}{3}. The gradient of the perpendicular line L2L_2 is 32-\frac{3}{2}, yielding the equation 3x+2y9=03x + 2y - 9 = 0. Setting y=0y = 0 gives the xx-intercept R(3,0)R(3, 0). Applying the internal section formula for ratio 1:21 : 2 on the xx-coordinates yields k+23=3\frac{k + 2}{3} = 3, which solves directly to k=7k = 7.

Adım Adım Çözüm

1
Find the midpoint MM of line segment ABAB and the gradient of line L1L_1.
Midpoint M=(2+42,1+52)=(1,3)M = \left(\frac{-2 + 4}{2}, \frac{1 + 5}{2}\right) = (1, 3). Gradient m1=514(2)=46=23m_1 = \frac{5 - 1}{4 - (-2)} = \frac{4}{6} = \frac{2}{3}.
Line L2L_2 passes through the midpoint MM and its orientation depends on the gradient of L1L_1.
2
Determine the gradient m2m_2 of L2L_2 and its equation.
Since L2L1L_2 \perp L_1, m2=1m1=32m_2 = -\frac{1}{m_1} = -\frac{3}{2}. Equation of L2L_2: y3=32(x1)    3x+2y9=0y - 3 = -\frac{3}{2}(x - 1) \implies 3x + 2y - 9 = 0.
Perpendicular lines have gradients whose product is 1-1 (m1m2=1m_1 \cdot m_2 = -1).
3
Find the coordinates of point RR, the xx-intercept of L2L_2.
Set y=0y = 0 in 3x+2y9=0    3x9=0    x=33x + 2y - 9 = 0 \implies 3x - 9 = 0 \implies x = 3. Thus, R=(3,0)R = (3, 0).
The xx-intercept occurs where y=0y = 0 on the coordinate plane.
4
Apply the section formula to find kk.
Point R(3,0)R(3, 0) divides P(1,2)P(1, -2) and Q(k,4)Q(k, 4) in ratio 1:21 : 2. The xx-coordinate is given by xR=1(k)+2(1)1+2=k+23x_R = \frac{1(k) + 2(1)}{1 + 2} = \frac{k + 2}{3}. Setting k+23=3\frac{k + 2}{3} = 3 gives k+2=9    k=7k + 2 = 9 \implies k = 7.
The internal section formula states that a point dividing (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) in ratio m:nm : n has coordinates (mx2+nx1m+n,my2+ny1m+n)\left(\frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n}\right).

Anahtar Kavram

Perpendicular lines, midpoints, intercepts, and section formula in coordinate geometry
Soru 7524Soru

A fair spinner is divided into 88 equal sectors numbered 11 to 88. In an experiment, the spinner is spun 120120 times, and it lands on a multiple of 33 a total of 3636 times. What is the positive difference between the experimental probability and the theoretical probability of landing on a multiple of 33?

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Cevap: 120\frac{1}{20}

Cevap

The positive difference between the experimental probability and theoretical probability is 120\frac{1}{20}.
The correct option is 120\frac{1}{20}. The theoretical probability of hitting a multiple of 33 on an 88-sector spinner is 28=14\frac{2}{8} = \frac{1}{4}. The experimental probability over 120120 spins with 3636 successes is 36120=310\frac{36}{120} = \frac{3}{10}. Subtracting theoretical probability from experimental probability gives 31014=6520=120\frac{3}{10} - \frac{1}{4} = \frac{6-5}{20} = \frac{1}{20}.

Adım Adım Çözüm

1
Calculate the theoretical probability of landing on a multiple of 33.
The sample space is {1,2,3,4,5,6,7,8}\{1, 2, 3, 4, 5, 6, 7, 8\} (total 88 outcomes). Multiples of 33 in this set are {3,6}\{3, 6\} (22 outcomes). Thus, P(Theoretical)=28=14P(\text{Theoretical}) = \frac{2}{8} = \frac{1}{4}.
Theoretical probability is the ratio of favorable outcomes to total possible outcomes in a fair trial.
2
Calculate the experimental probability (relative frequency) from the trial results.
P(Experimental)=36120=310P(\text{Experimental}) = \frac{36}{120} = \frac{3}{10}.
Experimental probability is defined as the number of observed successful events divided by the total number of trials conducted.
3
Find the positive difference between the experimental and theoretical probabilities.
Difference=31014=620520=120\text{Difference} = \frac{3}{10} - \frac{1}{4} = \frac{6}{20} - \frac{5}{20} = \frac{1}{20}.
Subtracting the smaller theoretical probability from the larger experimental probability yields the positive difference.

Anahtar Kavram

Experimental versus Theoretical Probability
Tahmini Süre:1m 30s
Soru 7525Soru

Two straight paths diverge from a junction JJ at an angle of 120120^\circ. A person walks 5 km5\text{ km} along the first path to point AA, and another person walks 16 km16\text{ km} along the second path to point BB. What is the direct distance between AA and BB in kilometres?

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Cevap: 19

Cevap

The direct distance between points AA and BB is 19 km19\text{ km}.
Using the Cosine Rule c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab \cos C with sides 5 km5\text{ km} and 16 km16\text{ km} and included angle 120120^\circ yields c2=52+1622(5)(16)(0.5)=25+256+80=361c^2 = 5^2 + 16^2 - 2(5)(16)(-0.5) = 25 + 256 + 80 = 361. Taking the square root gives 19 km19\text{ km}.

Adım Adım Çözüm

1
Identify given values and setup the Cosine Rule formula
Side a=5a = 5, side b=16b = 16, and included angle θ=120\theta = 120^\circ
The scenario provides two sides and the included angle (SAS configuration), which requires the Cosine Rule to find the third side.
2
Substitute the values into c2=a2+b22abcosθc^2 = a^2 + b^2 - 2ab \cos \theta
c2=52+1622(5)(16)cos(120)c^2 = 5^2 + 16^2 - 2(5)(16) \cos(120^\circ)
Populating the formula allows evaluation of the unknown distance squared.
3
Evaluate the trigonometric term and simplify
c2=25+256160(0.5)=281+80=361c^2 = 25 + 256 - 160(-0.5) = 281 + 80 = 361
The cosine of an obtuse angle in the second quadrant (120120^\circ) is negative: cos(120)=0.5\cos(120^\circ) = -0.5.
4
Take the square root to find the distance cc
c=361=19 kmc = \sqrt{361} = 19\text{ km}
Taking the positive square root gives the physical distance between the two points.

Anahtar Kavram

Applying the Cosine Rule to find the length of an unknown side in a non-right triangle given two sides and the included angle (SAS).
Soru 7526Soru

Find the indefinite integral (3x(x2)+8cos(4x))dx\int \left( 3x(x - 2) + 8\cos(4x) \right) dx.

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Cevap: x33x2+2sin(4x)+Cx^3 - 3x^2 + 2\sin(4x) + C

Cevap

x33x2+2sin(4x)+Cx^3 - 3x^2 + 2\sin(4x) + C
Expanding 3x(x2)3x(x - 2) yields 3x26x3x^2 - 6x. Integrating 3x23x^2 gives x3x^3, integrating 6x-6x yields 3x2-3x^2, and integrating 8cos(4x)8\cos(4x) yields 2sin(4x)2\sin(4x). Adding the arbitrary constant CC produces x33x2+2sin(4x)+Cx^3 - 3x^2 + 2\sin(4x) + C.

Adım Adım Çözüm

1
Expand the algebraic term inside the integrand.
3x(x2)=3x26x3x(x - 2) = 3x^2 - 6x, expressing the integral as (3x26x+8cos(4x))dx\int (3x^2 - 6x + 8\cos(4x)) dx.
Expanding the product converts the term into standard polynomial form suitable for integration.
2
Integrate each component term using basic integration rules.
3x2dx=x3\int 3x^2 dx = x^3, 6xdx=3x2\int -6x dx = -3x^2, and 8cos(4x)dx=8sin(4x)4=2sin(4x)\int 8\cos(4x) dx = 8 \cdot \frac{\sin(4x)}{4} = 2\sin(4x).
Apply the power rule xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} and trigonometric rule cos(kx)dx=sin(kx)k\int \cos(kx) dx = \frac{\sin(kx)}{k}.
3
Combine terms and attach the constant of integration.
x33x2+2sin(4x)+Cx^3 - 3x^2 + 2\sin(4x) + C
An indefinite integral requires an arbitrary constant CC to represent the full set of antiderivatives.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions
Soru 7527Soru

Evaluate the definite integral 03(x2+2)dx\int_{0}^{3} (x^2 + 2) \, dx.

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Cevap: 15

Cevap

The value of the definite integral is 15.
Integrating x2+2x^2 + 2 with respect to xx yields [x33+2x]03\left[\frac{x^3}{3} + 2x\right]_{0}^{3}. Substituting the upper bound x=3x = 3 gives 273+6=15\frac{27}{3} + 6 = 15. Substituting the lower bound x=0x = 0 gives 00. The net value is 150=1515 - 0 = 15.

Adım Adım Çözüm

1
Find the antiderivative of x2+2x^2 + 2
x33+2x\frac{x^3}{3} + 2x
Apply the power rule of integration to each term.
2
Substitute the upper limit x=3x = 3
15
\frac{3^3}{3} + 2(3) = 9 + 6 = 15
3
Substitute the lower limit x=0x = 0
0
033+2(0)=0\frac{0^3}{3} + 2(0) = 0
4
Compute the difference between upper and lower limit values
15
15 - 0 = 15

Anahtar Kavram

Definite Integration of Polynomial Functions
Tahmini Süre:45s
Soru 7528Soru

A bag contains 55 red balls, 33 blue balls, and nn green balls. The theoretical probability of drawing a blue ball at random from the bag is 320\frac{3}{20}. In a probability experiment, a ball is drawn at random from the bag and replaced 500500 times. If a green ball is observed 340340 times, calculate the absolute difference between the experimental probability and the theoretical probability of drawing a green ball.

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Cevap: 0.08

Cevap

The absolute difference between the experimental probability and the theoretical probability of drawing a green ball is 0.080.08 (or 225\frac{2}{25}).
The theoretical probability of drawing a blue ball establishes that the bag contains 2020 total balls, meaning there are 1212 green balls. The theoretical probability of picking a green ball is therefore 1220=0.60\frac{12}{20} = 0.60. The experimental probability is 340500=0.68\frac{340}{500} = 0.68. Taking the positive difference gives 0.680.60=0.08|0.68 - 0.60| = 0.08.

Adım Adım Çözüm

1
Find the total number of balls and the value of nn
n=12n = 12, Total balls = 2020
The theoretical probability of picking a blue ball is 38+n=320\frac{3}{8+n} = \frac{3}{20}, giving 8+n=208 + n = 20.
2
Find the theoretical probability of drawing a green ball
P(Green)theo=0.60P(\text{Green})_{\text{theo}} = 0.60
There are 1212 green balls out of 2020 total balls, so 1220=0.60\frac{12}{20} = 0.60.
3
Find the experimental probability of drawing a green ball
P(Green)exp=0.68P(\text{Green})_{\text{exp}} = 0.68
In 500500 trials, a green ball was observed 340340 times, so 340500=0.68\frac{340}{500} = 0.68.
4
Calculate the absolute difference
0.680.60=0.08|0.68 - 0.60| = 0.08
Subtract the theoretical probability from the experimental probability and take the absolute value.

Anahtar Kavram

Comparison of theoretical probability based on outcome sample space and experimental probability based on observed trial relative frequencies.
Soru 7529Soru

If y=extanx+ln(2x+1)y = e^{-x} \tan x + \ln(2x + 1), find the value of dydx\frac{dy}{dx} at x=0x = 0.

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Cevap: 3

Cevap

The value of dydx\frac{dy}{dx} at x=0x = 0 is 3.
Applying the product rule to extanxe^{-x}\tan x gives extanx+exsec2x-e^{-x}\tan x + e^{-x}\sec^2 x, and applying the chain rule to ln(2x+1)\ln(2x+1) gives 22x+1\frac{2}{2x+1}. Evaluating dydx=extanx+exsec2x+22x+1\frac{dy}{dx} = -e^{-x}\tan x + e^{-x}\sec^2 x + \frac{2}{2x+1} at x=0x = 0 yields e0(0)+e0(1)+21=3-e^0(0) + e^0(1) + \frac{2}{1} = 3.

Adım Adım Çözüm

1
Differentiate u(x)=extanxu(x) = e^{-x} \tan x using the product rule.
\frac{du}{dx} = -e^{-x} \tan x + e^{-x} \sec^2 x
The derivative of exe^{-x} is ex-e^{-x} and the derivative of tanx\tan x is \sec^2 x.
2
Differentiate v(x)=ln(2x+1)v(x) = \ln(2x + 1) using the chain rule.
dvdx=22x+1\frac{dv}{dx} = \frac{2}{2x + 1}
The derivative of ln(g(x))\ln(g(x)) is g(x)g(x)\frac{g'(x)}{g(x)}, where g(x)=2x+1g(x) = 2x + 1 and g(x)=2g'(x) = 2.
3
Sum the derivatives to find the complete expression for dydx\frac{dy}{dx}.
\frac{dy}{dx} = -e^{-x} \tan x + e^{-x} \sec^2 x + \frac{2}{2x + 1}
The derivative of a sum is equal to the sum of the derivatives.
4
Evaluate the derivative at x=0x = 0.
\frac{dy}{dx}\Big|_{x=0} = -e^0(0) + e^0(1)^2 + \frac{2}{1} = 3
Since tan(0)=0\tan(0) = 0, e0=1e^0 = 1, and sec(0)=1\sec(0) = 1, substituting x=0x = 0 simplifies the derivative to 0+1+2=30 + 1 + 2 = 3.

Anahtar Kavram

Differentiation of Transcendental Functions (Product and Chain Rules)
Soru 7530Soru

A constant-volume gas thermometer registers a pressure of 50kPa50\,\text{kPa} at the ice point (0C0^\circ\text{C}) and 70kPa70\,\text{kPa} at the steam point (100C100^\circ\text{C}). What is the temperature when the gas pressure measured by the thermometer is 62kPa62\,\text{kPa}?

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Cevap: 60C60^\circ\text{C}

Cevap

60C60^\circ\text{C}
The temperature θ\theta on the Celsius scale is given by the ratio of the change in thermometric property from the ice point to the total fundamental interval, scaled by 100. Substituting P0=50kPaP_0 = 50\,\text{kPa}, P100=70kPaP_{100} = 70\,\text{kPa}, and Pθ=62kPaP_\theta = 62\,\text{kPa} gives θ=62507050×100=1220×100=60C\theta = \frac{62 - 50}{70 - 50} \times 100 = \frac{12}{20} \times 100 = 60^\circ\text{C}.

Adım Adım Çözüm

1
Identify the given thermometric property values at the fixed points and target state
P0=50kPaP_0 = 50\,\text{kPa}, P100=70kPaP_{100} = 70\,\text{kPa}, and Pθ=62kPaP_\theta = 62\,\text{kPa}
These represent the lower fixed point, upper fixed point, and unknown temperature reading respectively.
2
Apply the general thermometric scale conversion formula
θ=PθP0P100P0×100C\theta = \frac{P_\theta - P_0}{P_{100} - P_0} \times 100^\circ\text{C}
Temperature on the Celsius scale varies linearly with the thermometric property relative to fixed points.
3
Substitute the values into the formula and solve for θ\theta
θ=62507050×100=1220×100=60C\theta = \frac{62 - 50}{70 - 50} \times 100 = \frac{12}{20} \times 100 = 60^\circ\text{C}
Carrying out the arithmetic yields the exact temperature of 60C60^\circ\text{C}.

Anahtar Kavram

Linear interpolation on temperature scales using thermometric properties
Tahmini Süre:1m 0s
Soru 7531Soru

A surveyor at station AA observes two landmarks, BB and CC. Landmark BB is located at a distance of 14 km14\text{ km} from AA on a bearing of 025025^\circ. Landmark CC is located at a distance of 62 km6\sqrt{2}\text{ km} from AA on a bearing of 070070^\circ. What is the direct distance between landmark BB and landmark CC?

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Cevap: 10 km10\text{ km}

Cevap

The direct distance between landmark BB and landmark CC is 10 km10\text{ km}.
The included angle BAC\angle BAC between the two bearings is 070025=45070^\circ - 025^\circ = 45^\circ. Applying the Cosine Rule a2=b2+c22bccosAa^2 = b^2 + c^2 - 2bc \cos A gives a2=(62)2+1422(62)(14)cos45=72+196168=100a^2 = (6\sqrt{2})^2 + 14^2 - 2(6\sqrt{2})(14)\cos 45^\circ = 72 + 196 - 168 = 100. Taking the square root gives 10 km10\text{ km}.

Adım Adım Çözüm

1
Determine the interior angle BAC\angle BAC from the given bearings.
BAC=070025=45\angle BAC = 070^\circ - 025^\circ = 45^\circ
The difference between two bearings measured clockwise from North from the same point gives the included angle between the lines of sight.
2
Identify the side lengths adjacent to angle AA in ΔABC\Delta ABC.
c=AB=14 kmc = AB = 14\text{ km}, b=AC=62 kmb = AC = 6\sqrt{2}\text{ km}, and included angle A=45A = 45^\circ
We have a Side-Angle-Side (SAS) triangle configuration, requiring the Cosine Rule to find the opposite side a=BCa = BC.
3
Apply the Cosine Rule a2=b2+c22bccosAa^2 = b^2 + c^2 - 2bc \cos A.
a2=(62)2+1422(62)(14)cos45a^2 = (6\sqrt{2})^2 + 14^2 - 2(6\sqrt{2})(14)\cos 45^\circ
Substituting known values into the Cosine Rule formula.
4
Simplify the terms and solve for aa.
a2=72+1961682(12)=268168=100    a=100=10 kma^2 = 72 + 196 - 168\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) = 268 - 168 = 100 \implies a = \sqrt{100} = 10\text{ km}
Squaring 626\sqrt{2} gives 36×2=7236 \times 2 = 72, 142=19614^2 = 196, and simplifying the cosine term yields 168168.

Anahtar Kavram

Cosine Rule for SAS non-right triangles in bearing contexts
Tahmini Süre:2m 0s
Soru 7532Soru

Find the value of xx, in degrees, for 0x900^\circ \le x \le 90^\circ that satisfies the trigonometric equation sin2x=cos(x+30)\sin 2x = \cos(x + 30^\circ).

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Cevap: 20

Cevap

The value of xx in the interval 0x900^\circ \le x \le 90^\circ satisfying the equation is 2020^\circ.
Using the co-function identity cosα=sin(90α)\cos \alpha = \sin(90^\circ - \alpha), we convert the right-hand side to sin(90(x+30))=sin(60x)\sin(90^\circ - (x + 30^\circ)) = \sin(60^\circ - x). Equating the arguments gives 2x=60x2x = 60^\circ - x, which simplifies to 3x=603x = 60^\circ, yielding x=20x = 20^\circ.

Adım Adım Çözüm

1
Apply the co-function trigonometric identity
cos(x+30)=sin(90(x+30))=sin(60x)\cos(x + 30^\circ) = \sin(90^\circ - (x + 30^\circ)) = \sin(60^\circ - x)
Converting cosine to sine allows direct comparison of sine functions on both sides of the equation.
2
Set up the equation equating the angle expressions
2x=60x2x = 60^\circ - x
Since sin(2x)=sin(60x)\sin(2x) = \sin(60^\circ - x) and x[0,90]x \in [0^\circ, 90^\circ], equating the principal angle arguments gives the primary solution.
3
Solve the linear equation for xx
3x=60    x=203x = 60^\circ \implies x = 20^\circ
Adding xx to both sides gives 3x=603x = 60^\circ, and dividing by 3 yields x=20x = 20^\circ.

Anahtar Kavram

Co-function identities and simple trigonometric equations
Soru 7533Soru

The mean of four positive integers is 99. If the mode of the set of numbers is 1212 and their median is 1010, what is the value of the smallest integer?

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Cevap: 44

Cevap

The smallest integer is 4.
Let the four positive integers in ascending order be abcda \le b \le c \le d. The sum of the four numbers is 4×9=364 \times 9 = 36. For 1212 to be the mode, it must appear at least twice, so c=12c = 12 and d=12d = 12. The median of four numbers is b+c2=10\frac{b + c}{2} = 10, which gives b+122=10\frac{b + 12}{2} = 10, so b=8b = 8. Substituting these into the sum equation a+8+12+12=36a + 8 + 12 + 12 = 36 gives a=4a = 4.

Adım Adım Çözüm

1
Calculate the sum of all four integers from the given mean.
Sum =4×9=36= 4 \times 9 = 36.
The mean of nn numbers is the sum divided by nn.
2
Determine the two largest numbers using the mode.
Let the ordered numbers be abcda \le b \le c \le d. Since the mode is 1212 and the set contains 44 numbers with a single mode, c=12c = 12 and d=12d = 12.
For 1212 to be the mode in a 4-element set without tie, it must appear at least twice.
3
Use the median to find the second integer bb.
Median =b+c2=b+122=10    b+12=20    b=8= \frac{b + c}{2} = \frac{b + 12}{2} = 10 \implies b + 12 = 20 \implies b = 8.
The median of an even number of ordered elements is the arithmetic mean of the two middle elements.
4
Solve for the smallest integer aa.
a+b+c+d=36    a+8+12+12=36    a+32=36    a=4a + b + c + d = 36 \implies a + 8 + 12 + 12 = 36 \implies a + 32 = 36 \implies a = 4.
Subtracting the sum of the known three numbers from the total sum gives the smallest integer.

Anahtar Kavram

Using mean, median, and mode definitions simultaneously to deduce unknown values in an ungrouped dataset
Soru 7534Soru

Two independent weather forecasting stations, AA and BB, operate in a region. The probability that station AA makes an accurate forecast on any given day is 0.800.80, and the probability that station BB makes an accurate forecast is 0.750.75. What is the probability that at least one of the two stations makes an accurate forecast on a given day?

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Cevap: 0.95

Cevap

The probability that at least one of the two stations makes an accurate forecast is 0.950.95.
The probability of at least one event occurring is given by P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B). Because events AA and BB are independent, P(AB)=P(A)×P(B)=0.80×0.75=0.60P(A \cap B) = P(A) \times P(B) = 0.80 \times 0.75 = 0.60. Substituting into the addition rule yields 0.80+0.750.60=0.950.80 + 0.75 - 0.60 = 0.95. Alternatively, using the complement rule: 1P(A)P(B)=1(10.80)(10.75)=1(0.20×0.25)=10.05=0.951 - P(A')P(B') = 1 - (1 - 0.80)(1 - 0.75) = 1 - (0.20 \times 0.25) = 1 - 0.05 = 0.95.

Adım Adım Çözüm

1
Calculate the probability of both events occurring simultaneously using the multiplication law for independent events.
P(AB)=P(A)×P(B)=0.80×0.75=0.60P(A \cap B) = P(A) \times P(B) = 0.80 \times 0.75 = 0.60
Since the two forecasting stations operate independently, the joint probability is the product of their individual probabilities.
2
Apply the general addition law of probability to calculate the probability of at least one station making an accurate forecast.
P(AB)=P(A)+P(B)P(AB)=0.80+0.750.60=0.95P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.80 + 0.75 - 0.60 = 0.95
The probability of compound event 'at least one' corresponds to the union of the two events.

Anahtar Kavram

Addition and Multiplication Laws of Probability for Independent Events
Soru 7535Soru

A transverse mechanical wave propagates along a stretched string. If the maximum speed of an oscillating particle of the string is equal to the speed of wave propagation, what is the ratio of the wave's amplitude to its wavelength?

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Cevap: 12π\frac{1}{2\pi}

Cevap

The ratio of the wave's amplitude to its wavelength is 12π\frac{1}{2\pi}.
For a transverse wave, the maximum speed of particle oscillation is vp,max=Aωv_{p,\text{max}} = A\omega, while the propagation speed of the wave pattern through the medium is vw=ωkv_w = \frac{\omega}{k}. Setting these two speeds equal gives Aω=ωkA\omega = \frac{\omega}{k}, which simplifies to Ak=1A k = 1. Substituting k=2πλk = \frac{2\pi}{\lambda} yields A(2πλ)=1A \left(\frac{2\pi}{\lambda}\right) = 1, giving the ratio Aλ=12π\frac{A}{\lambda} = \frac{1}{2\pi}.

Adım Adım Çözüm

1
Express the maximum speed of a particle in simple harmonic oscillation along a transverse wave.
The maximum particle speed is given by vp,max=Aωv_{p,\text{max}} = A\omega, where AA is the amplitude and ω\omega is the angular frequency.
Particles in a transverse mechanical wave oscillate perpendicularly to wave propagation in simple harmonic motion.
2
Express the wave propagation speed in terms of angular frequency and wave number.
The wave propagation speed is vw=ωkv_w = \frac{\omega}{k}, where k=2πλk = \frac{2\pi}{\lambda} is the angular wave number.
Wave propagation speed reflects the rate at which the wave energy and profile move through the medium.
3
Equate the maximum particle speed to the wave propagation speed and solve for Aλ\frac{A}{\lambda}.
Aω=ωk    Ak=1    A(2πλ)=1    Aλ=12πA\omega = \frac{\omega}{k} \implies A k = 1 \implies A \left(\frac{2\pi}{\lambda}\right) = 1 \implies \frac{A}{\lambda} = \frac{1}{2\pi}.
Setting vp,max=vwv_{p,\text{max}} = v_w allows the angular frequency ω\omega to cancel out, directly linking amplitude and wavelength.

Anahtar Kavram

Transverse Wave Particle Oscillation vs. Wave Propagation Speed
Tahmini Süre:2m 0s
Soru 7536Soru

What is the value of the definite integral 04xdx\int_{0}^{4} \sqrt{x} \, dx?

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Cevap: 163\frac{16}{3}

Cevap

163\frac{16}{3}
Applying the power rule to x1/2x^{1/2} gives 23x3/2\frac{2}{3}x^{3/2}. Substituting the upper limit x=4x = 4 yields 23(4)3/2=23(8)=163\frac{2}{3}(4)^{3/2} = \frac{2}{3}(8) = \frac{16}{3}, and substituting the lower limit x=0x = 0 yields 0. Subtracting the lower limit result from the upper limit result gives 163\frac{16}{3}.

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1
Express the integrand with a fractional exponent
x=x1/2\sqrt{x} = x^{1/2}
Rewriting the square root as a fractional exponent allows the application of the power rule of integration.
2
Find the antiderivative using the power rule of integration
x1/2dx=x1/2+11/2+1=x3/23/2=23x3/2\int x^{1/2} \, dx = \frac{x^{1/2 + 1}}{1/2 + 1} = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}
The power rule states that xndx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1} for n1n \neq -1.
3
Evaluate the antiderivative at the limits 4 and 0
[23x3/2]04=23(4)3/223(0)3/2=23(8)0=163\left[ \frac{2}{3}x^{3/2} \right]_{0}^{4} = \frac{2}{3}(4)^{3/2} - \frac{2}{3}(0)^{3/2} = \frac{2}{3}(8) - 0 = \frac{16}{3}
By the Fundamental Theorem of Calculus, evaluate F(b)F(a)F(b) - F(a).

Anahtar Kavram

Definite Integration using the Power Rule
Soru 7537Soru

For the domain 0x1800^\circ \le x \le 180^\circ, what is the complete solution set to the trigonometric equation 4sin2x3=04\sin^2 x - 3 = 0?

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Cevap: {60,120}\{60^\circ, 120^\circ\}

Cevap

{60,120}\{60^\circ, 120^\circ\}
Solving the equation 4sin2x3=04\sin^2 x - 3 = 0 gives sin2x=34\sin^2 x = \frac{3}{4}, which simplifies to sinx=32\sin x = \frac{\sqrt{3}}{2} within the interval 0x1800^\circ \le x \le 180^\circ. The principal angle is 6060^\circ. Since sine is positive in both the first and second quadrants, the second valid angle in the domain is 18060=120180^\circ - 60^\circ = 120^\circ, giving the solution set {60,120}\{60^\circ, 120^\circ\}.

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1
Isolate the trigonometric term
sin2x=34\sin^2 x = \frac{3}{4}
Rearrange 4sin2x3=04\sin^2 x - 3 = 0 by adding 3 to both sides and dividing by 4.
2
Take the square root of both sides
sinx=±32\sin x = \pm \frac{\sqrt{3}}{2}
Taking the square root yields both positive and negative ratios.
3
Apply the domain restriction 0x1800^\circ \le x \le 180^\circ
sinx=32\sin x = \frac{\sqrt{3}}{2}
The sine function is non-negative in the first and second quadrants (0x1800^\circ \le x \le 180^\circ), so the negative root has no solutions in this interval.
4
Determine the angles for xx
x = 60^\circ \text{ and } x = 180^\circ - 60^\circ = 120^\circ
The reference angle is 6060^\circ because sin60=32\sin 60^\circ = \frac{\sqrt{3}}{2}. In Quadrant II, the corresponding angle is 18060=120180^\circ - 60^\circ = 120^\circ.

Anahtar Kavram

Solving Quadratic Trigonometric Equations
Soru 7538Soru

What is the simplified numerical value of 123+1+1231\frac{\sqrt{12}}{\sqrt{3} + 1} + \frac{\sqrt{12}}{\sqrt{3} - 1}?

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Cevap: 6

Cevap

The simplified numerical value of the given expression is 6.
Combining the fractions using their common conjugate denominator (3+1)(31)=2(\sqrt{3}+1)(\sqrt{3}-1) = 2 leads to a numerator of 23[(31)+(3+1)]=23(23)=122\sqrt{3}[(\sqrt{3}-1)+(\sqrt{3}+1)] = 2\sqrt{3}(2\sqrt{3}) = 12. Dividing 12 by 2 yields the final answer of 6.

Adım Adım Çözüm

1
Simplify the surd in the numerator
\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}
Simplifying surds into basic form makes subsequent calculations simpler.
2
Combine the fractions by finding a common denominator
23(31)+23(3+1)(3+1)(31)\frac{2\sqrt{3}(\sqrt{3}-1) + 2\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)}
Multiplying denominators forms a conjugate pair, which rationalizes the combined denominator.
3
Simplify the numerator and the denominator
\text{Numerator} = 2\sqrt{3}(\sqrt{3}-1 + \sqrt{3}+1) = 2\sqrt{3}(2\sqrt{3}) = 12; \quad \text{Denominator} = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2
Expanding the numerator combines like surd terms, and using the difference of squares simplifies the denominator to a rational integer.
4
Perform the final division
122=6\frac{12}{2} = 6
Dividing the simplified numerator by the rationalized denominator gives the final integer value.

Anahtar Kavram

Rationalization of binomial denominators using conjugate surds
Soru 7539Soru

The sum of the first nn terms of a sequence is given by Sn=2n2+3nS_n = 2n^2 + 3n. Find the 5th5^{\text{th}} term of a geometric progression whose first term is the 3rd3^{\text{rd}} term of this sequence, and whose common ratio is equal to the common difference of this sequence.

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Cevap: 33283328

Cevap

The 5th term of the geometric progression is 3328.
Evaluating S1S_1, S2S_2, and S3S_3 gives sequence terms T1=5T_1 = 5, T2=9T_2 = 9, and T3=13T_3 = 13. The common difference is d=95=4d = 9 - 5 = 4. Using T3=13T_3 = 13 as the first term of the GP and r=4r = 4 as the common ratio, the 5th term of the GP is 13×451=13×256=332813 \times 4^{5-1} = 13 \times 256 = 3328.

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1
Calculate the first few terms of the sequence using the sum formula Sn=2n2+3nS_n = 2n^2 + 3n.
S1=2(1)2+3(1)=5S_1 = 2(1)^2 + 3(1) = 5, S2=2(2)2+3(2)=14S_2 = 2(2)^2 + 3(2) = 14, S3=2(3)2+3(3)=27S_3 = 2(3)^2 + 3(3) = 27.
The sum formula gives cumulative sums, from which individual terms can be derived.
2
Find the 3rd term (T3T_3) and the common difference (dd) of the arithmetic progression.
T1=5T_1 = 5, T2=S2S1=9T_2 = S_2 - S_1 = 9, T3=S3S2=13T_3 = S_3 - S_2 = 13. Common difference d=T2T1=4d = T_2 - T_1 = 4.
The difference between consecutive cumulative sums gives the individual sequence terms, and their constant difference gives the common difference.
3
Define the parameters of the geometric progression (GP).
First term of GP a=T3=13a = T_3 = 13, common ratio r=d=4r = d = 4.
The problem specifies that the first term of the GP is the 3rd term of the sequence and the common ratio equals the common difference.
4
Compute the 5th term of the geometric progression using Gn=arn1G_n = a \cdot r^{n-1}.
G5=13451=1344=13256=3328G_5 = 13 \cdot 4^{5-1} = 13 \cdot 4^4 = 13 \cdot 256 = 3328.
Applying the standard nth term formula for a GP with n=5n = 5.

Anahtar Kavram

Arithmetic Progression sum formula to term conversion and Geometric Progression nth term evaluation
Tahmini Süre:2m 0s
Soru 7540Soru

An arithmetic progression (A.P.) has a first term of 88 and a common difference of 66. What is the 9th9^{\text{th}} term of the progression?

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Cevap: 5656

Cevap

The 9th9^{\text{th}} term of the arithmetic progression is 5656.
The nthn^{\text{th}} term of an arithmetic progression is given by Tn=a+(n1)dT_n = a + (n - 1)d. Substituting a=8a = 8, d=6d = 6, and n=9n = 9 yields T9=8+(91)×6=8+48=56T_9 = 8 + (9 - 1) \times 6 = 8 + 48 = 56.

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1
Identify the given terms from the problem
First term a=8a = 8, common difference d=6d = 6, and term position n=9n = 9.
These are the parameters required for the nthn^{\text{th}} term formula of an A.P.
2
Apply the nthn^{\text{th}} term formula Tn=a+(n1)dT_n = a + (n - 1)d
T9=8+(91)×6T_9 = 8 + (9 - 1) \times 6
The nthn^{\text{th}} term requires multiplying the common difference by (n1)(n - 1).
3
Simplify the expression
T9=8+8×6=8+48=56T_9 = 8 + 8 \times 6 = 8 + 48 = 56
Perform multiplication before addition according to standard order of operations.

Anahtar Kavram

General term of an Arithmetic Progression
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