Find the? inverse, if it? exists, for the given matrix.

@MATX{{3;3;-1};{-12;-12;4};{2;6;0}}

Final answer:

To find the probability that a computer memory chip will have a lifetime of less than 8 years, we can use the properties of the Gamma distribution.

Explanation:

To find the probability that a computer memory chip will have a lifetime of less than 8 years, we can use the properties of the Gamma distribution. The average lifetime of the chip is given as 4 years, which corresponds to the mean. The variance is given as 16/3 years squared, which is equal to the mean squared.

Using these values, we can determine the shape and rate parameters of the Gamma distribution. The shape (α) is equal to the mean squared divided by the variance, which in this case is 16/3. The rate (β) is equal to the mean divided by the variance, which in this case is 4/(16/3).

To find the probability that the chip will have a lifetime of less than 8 years, we can calculate the cumulative distribution function (CDF) of the Gamma distribution with the shape and rate parameters we obtained.

Final answer:

The 95% confidence interval for the mean mathematics ACT score for all calculus students at this college is calculated to be between 24.693 and 27.307.

Explanation:

To calculate a 95% confidence interval for the mean mathematics ACT score for all calculus students at the college, we use the sample mean and standard deviation along with the z-score for a 95% confidence level. Since the sample size is large (n=81), we can use the z-distribution as an approximation for the t-distribution.

The formula for a confidence interval is:

Mean ± (z-score * (Standard Deviation / √n))
Here, the sample mean (μ) is 26, the standard deviation (s) is 6, and the sample size (n) is 81. For a 95% confidence level, the z-score is approximately 1.96.

Now we calculate the margin of error (ME):
ME = 1.96 * (6 / √81) = 1.96 * (6 / 9) = 1.96 * 0.6667 = 1.307

Therefore, the 95% confidence interval is
26 ± 1.307

Lower Limit = 26 - 1.307 = 24.693
Upper Limit = 26 + 1.307 = 27.307
The confidence interval is (24.693, 27.307).

We estimate with 95 percent confidence that the true population mean for the mathematics ACT score for all calculus students at the college is between 24.693 and 27.307.

Each calculation has been evaluated, taken all significant figures into consideration, and results have been presented in scientific notation. Special attention was given to rules related to multiplying numbers in scientific notation.

The given calculations require us to use scientific notation and proper treatment of significant figures. We are using the fundamentals of arithmetic with scientific notation, which is based on the rules of exponents. Each calculation is treated as follows:

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Answer:

a) 0; b) 4[tex]x^{2}[/tex]

Step-by-step explanation:

a) To compute f o g (0), first evaluate g(x) for x=0 and then evaluate f for x=g(0).

[tex]f \circ g (0)=f(2 \cdot 0)=f(0)=0^2[/tex]

b) To compute a mathematical expression for f o g do the same but instead of 0 use x,

[tex]f \circ g (x) = f( 2 \cdot x)= (2 \cdot x )^2[/tex]

In the question, f(x) = x², g(x) = 2x. We need to determine the value of function f composed with function g at 0 (f o g(0)), and the general expression for f o g(x). f o g(0) = 0 and (f o g)(x) = 4x².

To solve this problem, we first need to understand that 'f o g' denotes the composition of function f and function g, defined as (f o g)(x) = f(g(x)). In this case, function f(x) = x^2 and function g(x) = 2x.

(a) To evaluate f o g at 0, we substitute x = 0 into g(x), giving us g(0) = 2*0 = 0. Substituting g(0) into f(x), we get f(g(0)) = f(0) = 0. So, f o g(0) = 0.

(b) For a general form of f o g, we substitute g(x) = 2x into f(x), resulting in (f o g)(x) = f(2x) = (2x)^2 = 4x^2.

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Answer:

Given:  [tex]\begin{bmatrix}2&5&-2&1&2\\1&1&-2&1&4\end{bmatrix}[/tex]

[tex]2x_{1}+5 x_{2}  - 2_{3} + x_{4} + 2x_{5}\\[tex]

[tex]x_{1}+ x_{2}  - 2_{3} + x_{4} + 4x_{5}\\[/tex]

The system of linear equations in matrix form may be written as:

AX=B,

where,

A is coefficient matrix of order [tex]2\times 4[/tex] and is given by:

A = [tex]\begin{bmatrix}2&5&-2&1&2\\1&1&-2&1&4\end{bmatrix}[/tex]

X is variable matrix of order  [tex]5\times 1[/tex] and is given by:

X=  [tex]\begin{bmatrix}x_{_{1}}\\x_{_{2}}\\x_{3}\\x_{4}\\x_{5}\end{bmatrix}[/tex]

and B is the contant matrix of order [tex]2\times 1[/tex] and is given by:

B = [tex]\begin{bmatrix}1\\5\end{bmatrix}[/tex]

Now, AX=B

[tex]\begin{bmatrix}2&5&-2&1&2\\1&1&-2&1&4\end{bmatrix}[/tex]. [tex]\begin{bmatrix}x_{_{1}}\\x_{_{2}}\\x_{3}\\x_{4}\\x_{5}\end{bmatrix}[/tex] = [tex]\begin{bmatrix}1\\5\end{bmatrix}[/tex]

Answer:

(d) reliability

Step-by-step explanation:

mostly we see that  validity and reliability is the key aspects of all research they help in differentiation between good and bad research so both are very necessary aspects of any research so the another name for validity in quantity research is reliability.

so the reliability will be the correct answer

so option (d) will be correct option

Answer:

There are going to be 31 matches played in the soccer league.

Step-by-step explanation:

The soccer league has 6 teams, so if every team plays against the others twice, there are going to be played 30 matches:

-Team 1: v Team 2 (2), v Team 3 (2), v Team 4 (2), v Team 5 (2), v Team 6 (2)

-Team 2: v Team 3 (2), v Team 4 (2), v Team 5 (2), v Team 6 (2)

-Team 3: v Team 4 (2), v Team 5 (2), v Team 6 (2)

-Team 4: v Team 5 (2), v Team 6 (2)

-Team 5: v Team 6 (2)

-Team 6: -

If there is a final championship game after the 30 regular season matches, there are going to be 31 matches played in the league.

Answer:

Yes;Each side of triangle PQR is the same length as the corresponding side of triangle STU

Step-by-step explanation:

You can observe  the sides of both triangles to see if this property holds

Lets check the length of AB

A(0,3)   and B(0,-1)

[tex]AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\\\\\\AB=\sqrt{(0-0)^2+(-1-3)^2} \\\\\\\\AB=\sqrt{0^2+-4^2} \\\\\\AB=\sqrt{16} =4units[/tex]

Now check the length of the corresponding side DE

D(1,2)  and E(1,-2)

[tex]DE=\sqrt{(1-1)^2+(-2-2)^2} \\\\\\DE=\sqrt{0^2+-4^2} \\\\\\DE=\sqrt{16} =4units[/tex]

The side AB has the same length as side DE.This is also true for the remaining corresponding sides.

Answer:

The total number of samples that contain at least 3 red balls is 115.

Step-by-step explanation:

Total number of balls = 11

Total number of red balls = 6

Total number of white balls = 5

A sample of 4 balls is to be selected that contain at least 3 red. It means either 3 out of 4 balls are red or 4 out of 4 ball are red.

[tex]\text{Total ways}=\text{Three balls are red}+\text{Four balls are red}[/tex]

[tex]\text{Total ways}=^6C_3\times ^5C_1+^6C_4\times ^5C_0[/tex]

Combination formula:

[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

Using this formula we get

[tex]\text{Total ways}=\frac{6!}{3!(6-3)!}\times \frac{5!}{1!(5-1)!}+\frac{6!}{4!(6-4)!}\times \frac{5!}{0!(5-0)!}[/tex]

[tex]\text{Total ways}=20\times 5+15\times 1[/tex]

[tex]\text{Total ways}=115[/tex]

Therefore the total number of samples that contain at least 3 red balls is 115.

Using the combination formula, it is found that 115 samples contain at least 3 red balls.

The balls are chosen without replacement, which is why the combination formula is used.

Combination formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this problem, the outcomes with at least 3 red balls are:

Hence:

[tex]T = C_{6,3}C_{5,1} + C_{6,4} = \frac{6!}{3!3!}\frac{5!}{1!4!} + \frac{6!}{4!2!} = 20(5) + 15 = 100 + 15 = 115[/tex]

115 samples contain at least 3 red balls.

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Answer:

The answer is

[tex]W=\frac{S-2LH}{2H+2L}[/tex]

Step-by-step explanation:

The formula for the area of a solid rectangle is

[tex]S = 2HW+2LW+2LH[/tex]

Solve it for W

[tex]2HW+2LW=S-2LH\\\\W(2H+2L)=S-2LH\\\\W=\frac{S-2LH}{2H+2L}[/tex]

Answer:82,368

Step-by-step explanation:

Final answer:

The total number of eight-card hands formed from a standard 52-card deck where the hand consists of three cards of one denomination, three cards of another, and two cards of a third is 27,456. This is calculated by finding combinations of denominations and the specific cards within those denominations.

Explanation:

To answer the question of how many eight-card hands can be formed from a standard 52-card deck, where the hand consists of three cards of one denomination, three cards of another denomination, and two cards of a third denomination, we have to use combinations. This is a problem of combinatorics in which we are finding the number of ways to select items from a group without regard to order.

First, we choose the denominations. There are 13 denominations and we want to choose 3 of them for our hand. Using the combination formula, this can be done in C(13, 3) ways:

C(13, 3) = 13! / (3! × (13-3)!) = 286

After choosing the denominations, we need to select the specific cards. For each of the first two denominations, we select 3 out of the 4 available suits, and for the third denomination, we select 2 out of the 4 suits. This can be done as follows:

C(4, 3) for the first denomination: C(4, 3) = 4 ways
C(4, 3) for the second denomination: C(4, 3) = 4 ways
C(4, 2) for the third denomination: C(4, 2) = 6 ways

Multiplying the ways to choose the denominations by the ways to choose the cards for each denomination gives us the total number of distinct hands.

Total hands = C(13, 3) × C(4, 3) × C(4, 3) × C(4, 2) = 286 × 4 × 4 × 6 = 27,456

Therefore, there are 27,456 different eight-card hands that meet the given criteria.

Answer:

[tex](z-\frac{3}{2} )^2-\frac{29}{4}[/tex]

Step-by-step explanation:

We are given the following quadratic equation by completing the square:

[tex]z^2 - 3z - 5 = 0[/tex]

Rewriting the equation in the form [tex]x^2+2ax+a^2[/tex] to get:

[tex]z^2 - 3z - 5+(-\frac{3}{2} )^2-(-\frac{3}{2} )^2[/tex]

[tex]z^2-3z+(-\frac{3}{2} )^2=(z-\frac{3}{2} )^2[/tex]

Completing the square to get:

[tex] ( z - \frac{ 3 } { 2 } )^ 2 - 5 - ( - \frac { 3 } { 2 } ) ^ 2[/tex]

[tex](z-\frac{3}{2} )^2-\frac{29}{4}[/tex]

Answer: [tex]z_1=4.19\\\\z_2=-1.19[/tex]

Step-by-step explanation:

Add 5 to both sides of the equation:

[tex]z^2 - 3z - 5 +5= 0+5\\\\z^2 - 3z = 5[/tex]

Divide the coefficient of [tex]z[/tex] by two and square it:

[tex](\frac{b}{2})^2= (\frac{3}{2})^2[/tex]

Add it to both sides of the equation:

[tex]z^{2} -3z+ (\frac{3}{2})^2=5+ (\frac{3}{2})^2[/tex]

Then, simplifying:

[tex](z- \frac{3}{2})^2=\frac{29}{4}[/tex]

Apply square root to both sides and solve for "z":

[tex]\sqrt{(z- \frac{3}{2})^2}=\±\sqrt{\frac{29}{4} }\\\\z=\±\sqrt{\frac{29}{4}}+ \frac{3}{2}\\\\z_1=4.19\\\\z_2=-1.19[/tex]

Answer:

The probability that the racket is wood or defective is 0.57.

Step-by-step explanation:

Let W represents wood racket, G represents the graphite racket and D represents the defective racket,

Given,

n(W) = 100,

n(G) = 100,

⇒ Total rackets = 100 + 100 = 200

n(W∩D) = 15,

n(G∩D) = 14,

⇒ n(D) = n(W∩D) + n(G∩D) = 15 + 14 = 29,

We know that,

n(W∪D) = n(W) + n(D) - n(W∩D)

= 100 + 29 - 15

= 100 + 14

= 114,

Hence, the probability that the racket is wood or defective,

[tex]P(W\cup D) = \frac{114}{200}[/tex]

[tex]=0.57[/tex]

Answer: a) 48

b) 43

Step-by-step explanation:

Given : The number of students Professor Date has in his Calculus class = 31

The number of students Professor Date has in his Discrete Mathematics class = 17

(a) If we assume that there are no students who take both classes, then the total number of students Professor Date Has = 31+17=48

(b) If we assume that there are five students who take both classes, then the total number of students Professor Date Has = 31+17-5=43

By gcd, I think you mean gcf ( Greatest Common Factor).

To find the gcf find all the factors of each number:

Factors of 20: 1, 2, 4,5 ,10 , 20

Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56

The largest common factor is 4.

LCM = Least Common Multiple

This is the smallest number that both numbers divide into evenly

Find the prime factors of each number:

Prime factors of 20: 2 * 2 * 5

Prime factors of 56: 2 * 2 * 2 * 7

To find the LCM, multiply all prime factors the most number of times they occur.

In the prime factor of 56, 2 appears 3 times and 7 appears once.

In the prime factor of 20 5 appears once.

LCM = 2 * 2 *2 * 7 * 5 = 280

Final answer:

To calculate the population in the year 2010 based on exponential growth from 1990, use a growth rate factor and the known population figures from the given years.

Explanation:

Population Growth Calculation:

Determine the growth rate factor from 1990 to 1996: 25 million / 21.7 million = 1.152

Apply the growth rate to find the population in 2010: 21.7 million * (1.152)¹⁴ (14 years from 1996 to 2010) = 48.9 million

Answer:

[tex]a_{n}=15 + (-1)^n * 5[/tex]

Step-by-step explanation:

First, we notice that the when n is odd, [tex]a_{n}[/tex] = 10. And when n is even, [tex]a_{n}[/tex] = 20.

The average of 10 and 20 is [tex](10+20)/2 = 15[/tex]. So, the distance between 15 and 10 is the same that between 15 and 20.

That distance is 5.  

From 15, we need to subtract 5 to get 10 when n is odd and we need to add 5 to get 20 when n is even.

The easiest way to express that oscilation is using [tex](-1)^n[/tex], because it is (-1) when n is odd and 1 when is even. And when multiplied by 5, it will add or subtract 5 as we wanted.

Final answer:

To have the same number of plates, napkins, and cups, Whitney will need at least 80 of each.

Explanation:

To find the least number of packages of plates, napkins, and cups so that Whitney has the same number of each, we need to find the least common multiple (LCM) of 10, 16, and 8. The LCM is the smallest multiple that all three numbers have in common.

10 = 2 x 5, 16 = 2 x 2 x 2 x 2, and 8 = 2 x 2 x 2

We can identify the prime factors of each number and then multiply the highest factor from each number. In this case, the LCM is 2 x 2 x 2 x 2 x 5 = 80.

Therefore, Whitney will need at least 80 plates, 80 napkins, and 80 cups to have the same number of each.

Answer:

Expected Grade=2 i.e., C

Variance=1.2

Step-by-step explanation:

[tex]Expected\ value=E\left [ x \right ]=\sum _{i=1}^{k} x_{i}p_{i}[/tex]

The x values are

A = 4

B = 3

C = 2

D = 1

F = 0

Probability of each of the events

P(4)=0.1

P(3)=0.2

P(2)=0.4

P(1)=0.2

P(0)=0.1

[tex]E\left [ x \right ]=4\times 0.1+3\times 0.2+2\times 0.4+1\times 0.2+0\times 0.1\\\therefore E\left [ x \right ]=2[/tex]

Variance

[tex]Var\left ( x\right)=E\left [ x^2 \right ]-E\left [ x \right ]^2[/tex]

[tex]E\left [ x^2 \right ]=4^2 \times 0.1+3^2 \times 0.2+2^2 \times 0.4+1^2 \times 0.2+0^2 \times 0.1\\\Rightarrow E\left [ x^2 \right ]=5.2\\E\left [ x \right ]^2=2^2=4\\\therefore Var\left ( x\right)=5.2-4=1.2\\[/tex]

Answer:

sin Ф=3/√13

Cos Ф=2/√13

Tan Ф=3/2

Step-by-step explanation:

Let x=2

Let y=3

Let r be the length of line segment drawn from origin to the point

[tex]r=\sqrt{x^2+y^2}[/tex]

Find r

[tex]r=\sqrt{2^2+3^2} =\sqrt{4+9} =\sqrt{13}[/tex]

Apply the relationship for sine, cosine and tan of Ф where

r=hypotenuse

Sine Ф=length of opposite side÷hypotenuse

Sin Ф=O/H where o=3, hypotenuse =√13

sin Ф=3/√13

CosineФ=length of adjacent side÷hypotenuse

Cos Ф=A/H

Cos Ф=2/√13

Tan Ф=opposite length÷adjacent length

TanФ=O/A

Tan Ф=3/2

Answer:

8000/4*100 = $200'000

Step-by-step explanation:

Answer:

$1.710

Step-by-step explanation:

Mike deposited $850 into the bank in July.

In August his balance will be: $850×1.15 = $977.5

In September his balance will be: $977.5×1.15 = $1124.125

In October his balance will be: $1124.125×1.15 = $1292.74375

In November his balance will be: $1292.74375×1.15 = $1.486,6553125

In December his balance will be: $1.486,6553125×1.15 = $1.709,653609375

Therefore, the amount of money he will have after december will be $1.710

Answer:

  90 adults; 160 students

Step-by-step explanation:

Let "a" represent the number of adults who went. The number of students can be represented by (250-a). Then the total cost of tickets is ...

  4.50a +2.50(250-a) = 805

  2a + 625 = 805 . . . . . . simplify

  2a = 180 . . . . . . . . . . . . subtract 625

  a = 90 . . . . . . . . . . . . . . divide by 2

  # of students = 250 -90 = 160

160 students and 90 adults went to the movies.

Answer:

the answer is 90 adults , and 160 students

Answer:

(141)₁₀→(241)₇

Step-by-step explanation:

(141)₁₀→(?)₇

for conversion of number from decimal to base 7 value we have to

factor 141 by 7

which is shown in the figure attached below.

from the attached figure we can clearly see that the colored digit will

give the conversion

we will write the digit from the bottom as shown in figure

(141)₁₀→(241)₇

Answer:

a) $ 1465.418

b) $ 214.582

Step-by-step explanation:

Since, the monthly payment formula of a loan is,

[tex]P=\frac{PV(r)}{1-(1+r)^{-n}}[/tex]

Where, PV is the principal amount of the loan,

r is the monthly rate,

n is the total number of months,

Here, P = $ 30, r = 1.25 % = 0.0125, n = 36 ( since, time is 3 years also 1 year = 12 months )

Substituting the values,

[tex]30=\frac{PV(0.0125)}{1-(1+0.0125)^{-36}}[/tex]

By the graphing calculator,

[tex]PV=865.418[/tex]

a) Thus, the cost of the TV = Down Payment + Principal value of the loan

= $ 600 + $ 865.418

= $ 1465.418

b) Now, the total payment = Monthly payment × total months

= 30 × 36

= $ 1080

Hence, the total amount of interest paid = total payment - principal value of the loan

=  $ 1080 - 865.418

=  $ 214.582.

Answer:

The height is 28.57 cm.

The surface area is 9,628 cm^2.

Step-by-step explanation:

I assume the cooler is shaped like a rectangular prism with length and width of the base given, and with an unknown height.

volume = length * width * height

First, we convert the volume from liters to cubic centimeters.

60 liters * 1000 mL/L * 1 cm^3/mL = 60,000 cm^3

Now we substitute every dimension we have in the formula and solve for height, h.

60,000 cm^3 = 60 cm * 35 cm * h

60,000 cm^3 = 2,100 cm^2 * h

h = (60,000 cm^3)/(2,100 cm^2)

h = 28.57 cm

The height is 28.57 cm.

Now we calculate the internal surface area.

total surface area = area of the bases + area of the 4 sides

SA = 2 * 60 cm * 35 cm + (60 cm + 35 cm + 60 cm + 35 cm) * 28.57 cm

SA = 9,628 cm^2

The surface area is 9,628 cm^2.

Explanation:

This can be explained by thinking numbers on the number line as:

Lets take we have to multiply a positive number (say, 2) with a negative number say (-3)

2×(-3)

Suppose someone is standing at 0 on the number line and to go to cover -3 , the person moves 3 units in the left hand side. Since, we have to compute for 2×(-3), The person has to cover the same distance twice. At last, he will be standing at -6, which is a negative number.

A image is shown below to represent the same.

Thus, a positive times a negative is a negative number.

Answer:[tex]\frac{1}{1000}[/tex]

Step-by-step explanation:

For the lottery game three numbers must match in exact order

From 0 to 9 total 10 numbers are there

Therefore selecting exactly same numbers as of winner is

=[tex]^{10}C_1\times ^{10}C_1\times ^{10}C_1 [/tex]

Since numbers are repeatable therefore each time we have a choice of choosing 1 number out o 10

=[tex]10\times 10\times 10[/tex]

Probability of winning=[tex]\frac{1}{1000}[/tex]

In a lottery game where the player must match a repeatable sequence of three numbers ranging from 0-9, the probability of a single ticket having the winning sequence is 1/1000, or 0.1%.

The subject of this problem is probability in Mathematics, specifically in a lottery context. Each digit in the sequence can be any number from 0 to 9. Because these numbers are repeatable, this means there are 10 possible numbers for each of the three digits in the sequence. Therefore, to find the total number of possible sequences, you multiply the ten options for the first digit by the ten for the second and the ten for the third, which comes to 10 * 10 * 10 = 1000 total possible sequences.

Since you are only looking for one specific sequence being the winning number, that means there’s only 1 favorable outcome out of 1000. Therefore, the probability of your ticket having the winning number sequence is 1/1000, or 0.001 in decimal form or 0.1% in percentage form.

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Answer:

Bob performed better in mathematics part of the SAT than the ACT

Step-by-step explanation:

We need to calculate the z-scores for both parts and compare them.

Z-score for the SAT is calculated using the formula:

[tex]Z=\frac{X-\mu}{\sigma}[/tex]

where [tex]\mu=514[/tex] is the mean and [tex]\sigma=113[/tex] is the standard deviation, and [tex]X=660[/tex] is the SAT test score.

We plug in these values to obtain:

[tex]Z=\frac{660-514}{113}[/tex]

[tex]Z=\frac{146}{113}=1.29[/tex] to the nearest hundredth.

We use the same formula to calculate the z-score for the ACT too.

Where [tex]\mu=20.6[/tex] is the mean and [tex]\sigma=5.1[/tex] is the standard deviation, and [tex]X=27[/tex] is the ACT test score.

We substitute the values to get:

[tex]Z=\frac{27-20.6}{5.1}=1.25[/tex] to the nearest hundredth.

Since 1.29 > 1.25, Bob performed better in mathematics part of the SAT

Using z-scores to determine where Bob's performance stands compared to others, we find that he performed slightly better on the SAT with a z-score of 1.29, than on the ACT with a z-score of 1.25.

To determine on which test Bob performed better, we have to calculate his z-scores on both the SAT and ACT. Z-score is a statistical measurement that describes a score's relationship to the mean of a group of scores. It indicates how many standard deviations an element is from the mean.

Here is how you calculate the z-score: z = (X - μ) / σ, where X is the individual score, μ is the mean, and σ is the standard deviation.

We can apply this formula to both of Bob's scores. For the SAT: z = (660 - 514) / 113 ≈ 1.29 For the ACT: z = (27 - 20.6) / 5.1 ≈ 1.25

Comparing the z-scores, Bob's score is above the mean by 1.29 standard deviations on the SAT and by 1.25 standard deviations on the ACT. Therefore, Bob performed slightly better on the SAT than on the ACT when comparing his scores to other test takers.

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Answer:

Suppose,

(C - A) ∩ (B - A) ≠ ∅

Let x is an element of (C - A) ∩ (B - A),

That is, x ∈ (C - A) ∩ (B - A),

⇒ x ∈ C - A and x ∈ B - A

⇒ x ∈ C, x ∉ A and x ∈ B, x ∉ A

⇒ x ∈ B ∩ C and x ∉ A

⇒ B ∩ C ⊄ A

But we have given,

B ∩ C ⊂ A

Therefore, our assumption is wrong,

And, there is no common elements in (C - A) and (B-A),

That is,  (C - A) ∩ (B - A) = ∅

Hence proved...