The cartesian product of two sets P and Q can be written as

Answer:

The part of the bridge that opens is 50 ft.

Step-by-step explanation:

The given parameters of the drawbridge are;

The entire length of the bridge = 100 feet

The height of the isosceles trapezoid formed = 25 feet

The angle at which the drawbridge meets the land ≈ 60°

Therefore, the part of the bridge that opens = The top narrow parallel side of the isosceles trapezoid

The length of each half of the bridge = (The entire length)/2 = 100 ft./2 = 50 ft.

Let 'x' represent the path of the waterway still partly blocked by each half of the bridge inclined

∴ x = 50 × cos(60°) = 25

x = 25 ft.

The path covered by both sides of the drawbridge = 2·x = 2 × 25 ft. = 50 ft.

The part of the bridge that opens = The entire length - 2·x

∴ The part of the bridge that opens = 100 ft. - 50 ft. = 50 ft.

The part of the bridge that opens = 50 ft.

Answer:

Step-by-step explanation:

Means of means = means of extremes 8y = x

x = 8y

Option B is the correct answer

Answer:

Surface area of cuboid = 78 unit²

Step-by-step explanation:

Given diagram is a cuboid prism

Given:

Length of cuboid = 5 unit

Width of cuboid = 3 unit

Height of cuboid = 3 unit

Find:

Surface area of cuboid

Computation:

Surface area of cuboid = 2[lb + bh + hl]

Surface area of cuboid = 2[(5)(3) + (3)(3) + (3)(5)]

Surface area of cuboid = 2[15 + 9 + 15]

Surface area of cuboid = 2[39]

Surface area of cuboid = 78 unit²

Answer: p+(-q)

Step-by-step explanation:

Full question:

Doug's teacher told him that the standardized score (z-score) for his mathematics exam, as compared to the exam scores of other students in the course, is 1.20. Which of the following is the best interpretation of this standardized score?

Doug's test score is 120.

Doug's test score is 1.20 times the average test score of students in the course.

Doug's test score is 1.20 above the average test score of students in the course.

Doug's test score is 1.20 standard deviations above the average test score of students in the course.

None of the above gives the correct interpretation.

Answer:

Doug's test score is 1.20 standard deviations above the average test score of students in the course.

Explanation:

Z scores are also known as standardized scores or normal scores or standardized variables. Z scores are used to standardize raw data in order to give them a uniformity or standard that allows for easier comparison of data values. For us to calculate a z-score as was done in Doug's test score, we simply subtract the mean from the raw data score and we divide the answer by the standard deviation.

The 90% confidence interval for the given sample is (7.58, 8.60).

To calculate the 90% confidence interval for the given sample assuming normality of the data, we need to use the formula as follows;Confidence interval = X ± Z α/2(σ/√n)Where, X is the sample meanZ α/2 is the Z-score for the desired level of confidenceσ is the population standard deviationn is the sample sizeFirst, we need to calculate the sample mean and standard deviation.Sample mean,

X= (7.9 + 8.3 + 8.4 + 9.6 + 7.7 + 8.1 + 6.8 + 7.5 + 8.6 + 8 + 7.8 + 7.4 + 8.4 + 8.9 + 8.5 + 9.4 + 6.9 + 7.7) / 18

= 8.09

Sample standard deviation,

σ = √[Σ(xi - X)² / (n - 1)]σ = √[(7.9 - 8.09)² + (8.3 - 8.09)² + (8.4 - 8.09)² + (9.6 - 8.09)² + (7.7 - 8.09)² + (8.1 - 8.09)² + (6.8 - 8.09)² + (7.5 - 8.09)² + (8.6 - 8.09)² + (8 - 8.09)² + (7.8 - 8.09)² + (7.4 - 8.09)² + (8.4 - 8.09)² + (8.9 - 8.09)² + (8.5 - 8.09)² + (9.4 - 8.09)² + (6.9 - 8.09)² + (7.7 - 8.09)² / (18 - 1)]σ = 0.761

Now, we need to find the Z α/2 value from the standard normal distribution table.

Z α/2 = 1.645 (for 90% confidence level)Putting the values in the formula,Confidence interval =

X ± Z α/2(σ/√n)

= 8.09 ± 1.645(0.761/√18)

= 8.09 ± 0.511

= (8.09 - 0.511, 8.09 + 0.511)

= (7.58, 8.60).

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Answer:

The area of the parallelogram is 20cm²

Answer:

A = 20 cm²

Step-by-step explanation:

The area (A) of a parallelogram is calculated as

A = bh ( b is the base and h the height ) , then

A = 5 × 4 = 20 cm²

Answer:

250/5 =50 miles per hour

Answer:

Buddy this might not be the correct answer but I got either 98 or 418 inches. Don't quote me on it though.

Step-by-step explanation:

The approximate interest rate needed to turn $3,500 into $7,000 over 10 years is 9%. Correct answer is C.

The value of money increases over time with the help of compounding interest. If one puts in a principal amount in an account, the amount will increase over time as interest accrues. Let's use the future value formula for the calculation. Let’s assume that the interest rate needed to turn $3,500 into $7,000 over 10 years is x percent. P = $3,500 (principal)FV = $7,000 (future value)

N = 10 years (duration of the investment)Using the future value formula:

FV = P(1 + r/n)^(nt)where, r is the annual interest rate, n is the number of times the interest is compounded in a year, and t is the duration of the investment in years.

Substituting the given values, we have: $7,000 = $3,500(1 + x/n)^(n × 10)We can solve for x by approximating the interest rate using each of the answer options given in the question until we find an answer that is close to $7,000. A calculator can also be used to calculate the compound interest for each option. If the interest rate is 7%, then the interest is compounded annually. Therefore, n = 1$7,000

= $3,500(1 + 0.07/1)^(1 × 10) If the interest rate is 10%, then the interest is compounded annually.

Therefore, n = 1$7,000 = $3,500(1 + 0.1/1)^(1 × 10)Thus, x ≈ 9.57%, greater than the required amount.

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Answer:

g= 2x+9 /3x

Step-by-step explanation:

By applying the Runge-Kutta method with a step size (h) of 0.09, we can estimate the value of the solution at t = 0.1 for the differential equation y' = 3 + t - y, with the initial condition y(0) = 1.

The Runge-Kutta method is a numerical technique used to approximate the solution of ordinary differential equations. In this case, we have the differential equation y' = 3 + t - y, where y' represents the derivative of y with respect to t. To apply the Runge-Kutta method, we need to iterate through the given range of t values, which is from 0 to 0.1 in this case, with a step size (h) of 0.09.

We start with the initial condition y(0) = 1. Then, for each iteration, we calculate the slope at the current point using the given equation. Using the slope, we estimate the value of y at the next time step (t + h). This process is repeated until we reach the desired value of t = 0.1.

By applying the Runge-Kutta method with h = 0.09, we can obtain an estimate for the value of y at t = 0.1. This method provides a more accurate approximation compared to simpler methods like Euler's method, as it considers multiple intermediate steps to improve accuracy.

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Answer:

3 years approx

Step-by-step explanation:

Given data

Principal=£100

Amount= £109.27

Rate= 3%

The expression for the compound interest is

A=P(1+r)^t

Make t subject of formula we have

t= ln(A/P) / r

t= ln(109.27/100)/ 3

t= ln(1.0927)/0.03

t= 0.088/0.03

t= 2.93

Hence the time is 3 years approx

A. The mean ÿ(t) = 0 and the autocorrelation Ry(t + 7, t) = 4e⁻⁷. Y(t) is wide-sense stationary.

B. the cross-correlation Rxy(t + 7, t) = 2e⁻⁷. X and Y are jointly wide-sense stationary.

C. The autocovariance Cy(t + 7, t) = 4e⁻⁷. Y(t) is not a white process because autocovariance Cy(t + 7, t) is not a Dirac delta function.

To find the mean ÿ(t) = E[Y(t)] and the autocorrelation Ry(t + 7, t) = E[Y(t + 7)Y(t)], we substitute the expression for Y(t) into the formulas:

(a) Mean of Y(t):

ÿ(t) = E[Y(t)] = E[2X(t) + sin(2t)]

= 2E[X(t)] + E[sin(2t)]

= 2(0) + 0

= 0

(b) Autocorrelation of Y(t + 7, t):

Ry(t + 7, t) = E[Y(t + 7)Y(t)]

= E[(2X(t + 7) + sin(2(t + 7)))(2X(t) + sin(2t))]

Expanding the expression:

Ry(t + 7, t) = E[4X(t + 7)X(t) + 2X(t + 7)sin(2t) + 2sin(2(t + 7))X(t) + sin(2(t + 7))sin(2t)]

Since X(t) is a WSS random process with mean 0, its autocorrelation Rx(T) = E[X(t + 7)X(t)] = e^(-|7|).

Using the properties of expectation and the independence of X(t) and sin(2t):

Ry(t + 7, t) = 4E[X(t + 7)X(t)] + 2E[X(t + 7)]E[sin(2t)] + 2E[sin(2(t + 7))]E[X(t)] + E[sin(2(t + 7))]E[sin(2t)]

= 4Rx(7) + 2(0)(0) + 2(0)(0) + 0

= 4e⁻⁷

Therefore, the mean ÿ(t) = 0 and the autocorrelation Ry(t + 7, t) = 4e⁻⁷.

To determine if Y(t) is wide-sense stationary, we need to check if the mean and autocorrelation are independent of time:

Mean: The mean ÿ(t) is constant and does not depend on time t. Thus, Y(t) has a constant mean.

Autocorrelation: The autocorrelation Ry(t + 7, t) depends only on the time difference of 7. It is independent of the absolute values of t. Therefore, Y(t) has a stationary autocorrelation.

Since Y(t) has a constant mean and a stationary autocorrelation, it is wide-sense stationary.

Moving on to part (b), we need to find the cross-correlation Rxy(t + 7, t) = E[X(t + 7)Y(t)].

Rxy(t + 7, t) = E[X(t + 7)Y(t)]

= E[X(t + 7)(2X(t) + sin(2t))]

Expanding the expression:

Rxy(t + 7, t) = E[2X(t + 7)X(t) + X(t + 7)sin(2t)]

Since X(t) is a WSS random process, its autocorrelation Rx(T) = e|⁻⁷|.

Using the properties of expectation and the independence of X(t) and sin(2t):

Rxy(t + 7, t) = 2E[X(t + 7)X(t)] + E[X(t + 7)]E[sin

(2t)]

= 2Rx(7) + 0

= 2e⁻⁷

Therefore, the cross-correlation Rxy(t + 7, t) = 2e⁻⁷.

To determine if X and Y are jointly wide-sense stationary, we need to check if the cross-correlation Rxy(t + 7, t) is independent of time:

Cross-correlation: The cross-correlation Rxy(t + 7, t) depends only on the time difference of 7. It is independent of the absolute values of t. Therefore, X and Y have a stationary cross-correlation.

Since the cross-correlation is stationary, X and Y are jointly wide-sense stationary.

Moving on to part (c), we need to find the autocovariance Cy(t + 7, t) = E[(Y(t + 7) - ÿ(t + 7))(Y(t) - ÿ(t))].

Expanding the expression:

Cy(t + 7, t) = E[(2X(t + 7) + sin(2(t + 7))) - 0][(2X(t) + sin(2t)) - 0]

= E[(2X(t + 7) + sin(2(t + 7)))(2X(t) + sin(2t))]

Using the same approach as in part (b), we expand the expression and evaluate the expectation:

Cy(t + 7, t) = 4E[X(t + 7)X(t)] + 2E[X(t + 7)]E[sin(2t)] + 2E[sin(2(t + 7))]E[X(t)] + E[sin(2(t + 7))]E[sin(2t)]

= 4Rx(7) + 0 + 0 + 0

= 4e⁻⁷

Therefore, the autocovariance Cy(t + 7, t) = 4e⁻⁷.

To determine if Y(t) is white, we check if the autocovariance Cy(t + 7, t) is a Dirac delta function. Since Cy(t + 7, t) = 4e⁻⁷ ≠ 0, it is not a Dirac delta function. Hence, Y(t) is not a white process.

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The solution to the given initial value problem using the method of Laplace transforms, is: y(t) = -4 [tex]e^{-t}[/tex] + 5 [tex]e^{-3t/5}[/tex]

To solve the given initial value problem using the method of Laplace transforms, we will follow these steps:

Taking the Laplace transform of both sides of the differential equation.

Applying the Laplace transform to the given differential equation, we get:

5L{y''} + 2L{y'} + 3L{y} = L{u(t-[tex]\pi[/tex])}

Using the properties of Laplace transforms and the table of Laplace transforms to simplify the equation.

The Laplace transform of y'' is [tex]s^2[/tex]Y(s) - sy(0) - y'(0), where Y(s) is the Laplace transform of y(t).

The Laplace transform of y' is sY(s) - y(0), and the Laplace transform of y is Y(s).

Using these transformations and considering the initial conditions y(0) = 1 and y'(0) = 1, we can rewrite the equation as:

5([tex]s^2[/tex]Y(s) - s - 1) + 2(sY(s) - 1) + 3Y(s) = e^(-pi*s) / s

Simplifying further, we have:

(5[tex]s^2[/tex] + 2s + 3)Y(s) - (5s + 7) = [tex]e^{-\pi s}[/tex] / s

Solving for Y(s):

Rearranging the equation, we get:

Y(s) = ([tex]e^{-\pi s}[/tex] / s + (5s + 7)) / (5[tex]s^2[/tex] + 2s + 3)

Using partial fraction decomposition to express Y(s) in simpler terms.

Performing partial fraction decomposition on the right side, we can express Y(s) as:

Y(s) = A / (s + 1) + B / (5s + 3)

where A and B are constants to be determined.

Using the inverse Laplace transform, we can find the solution y(t) as:

y(t) = [tex]L^{-1}[/tex]{Y(s)} = [tex]L^{-1}[/tex]{A / (s + 1)} + [tex]L^{-1}[/tex]{B / (5s + 3)}

Taking the inverse Laplace transforms using the table of Laplace transforms, we find:

y(t) = A [tex]e^{-t}[/tex] + B [tex]e^{-3t/5}[/tex]

Substituting the initial conditions y(0) = 1 and y'(0) = 1 into the solution y(t) = A [tex]e^{-t}[/tex] + B [tex]e^{-3t/5}[/tex], we can solve for the constants A and B.

First, substitute t = 0 into the equation:

y(0) = A * [tex]e^{-0}[/tex] + B * [tex]e^{-0}[/tex] = A + B = 1

Next, differentiate the solution y(t) with respect to t:

y'(t) = -A * [tex]e^{-t}[/tex] - (3B/5) * [tex]e^{-3t/5}[/tex]

Then, substitute t = 0 and y'(0) = 1 into the equation:

y'(0) = -A * [tex]e^{-0}[/tex] - (3B/5) * [tex]e^{-0}[/tex] = -A - (3B/5) = 1

We now have a system of equations:

A + B = 1

-A - (3B/5) = 1

Solving this system of equations, we can find the values of A and B.

From the first equation, we can rewrite it as:

A = 1 - B

Substituting this expression for A into the second equation:

-(1 - B) - (3B/5) = 1

Simplifying the equation:

-1 + B - (3B/5) = 1

Multiplying through by 5 to eliminate the fraction:

-5 + 5B - 3B = 5

Combining like terms:

2B = 10

Dividing by 2:

B = 5

Substituting the value of B back into the first equation:

A = 1 - 5 = -4

Therefore, the constants A and B are -4 and 5, respectively.

The solution to the initial value problem is:

y(t) = -4 [tex]e^{-t}[/tex] + 5 [tex]e^{-3t/5}[/tex]

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The solution to the recurrence relation an = 3an-1 + 4an-2 with initial values ao = 2 and a₁ = 3 is given byan = (-1)4ⁿ - 4(4)ⁿ-¹/16

Given recurrence relation is an = 3an-1 + 4an-2, with initial values ao = 2 and a₁ = 3.

The characteristic equation of the recurrence relation is given byr² - 3r - 4 = 0

Solving the characteristic equation, we get

r² - 4r + r - 4 = 0

r(r - 4) + 1(r - 4) = 0

(r - 4)(r + 1) = 0

r1 = 4, r2 = -1

So, the general solution of the recurrence relation is given by

an = Ar¹ + Br²

For r1 = 4, a4 = 3

a3 + 4a2a4 = 3a3 + 4a2 = 3(4a2 + 4a1) + 4a2= 16a2 + 12a1 ....(1)

For r2 = -1, aₙ₊₁ = 3an + 4an-1aₙ₊₁ = 3an + 4an-1 = 3(A(-1)^n + B(4)^n) + 4(A(-1)^(n-1) + B(4)^(n-1))= 3A(-1)^n - 4A(-1)^(n-1) + 12B(4)^n + 4B(4)^(n-1)= A(-1)^n + 4B(4)^n ....(2)

Putting n = 0 in (2), we get

a1 = A - 4A = -3A = 3 => A = -1

Substituting A = -1 in (1), we get

a4 = 16a2 + 12a1=> a4 = 16a2 + 12(2) => a4 = 16a2 + 24a4 = 16a2 + 24 => a2 = (a4 - 24)/16

Thus the solution to the recurrence relation an = 3an-1 + 4an-2 with initial values ao = 2 and a₁ = 3 is given by

an = (-1)4ⁿ - 4(4)ⁿ-¹/16

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Answer: 3 times greater

Step-by-step explanation:

Height: 4x3=12

Length: 3x3=9

Width: 3x3=9

The sides on the red cuboid times by 3 equals the sides on the purple one.

Hope this helps :)

Answer:

10 miles

Step-by-step explanation:

The information given forms a right angled triangle ; hence, we can use Pythagoras rule to solve for the distance, x

Recall:

Hypotenus = sqrt(opposite ² + adjacent ²)

Hypotenus = x

Therefore,

x = sqrt(6² + 8²)

x = sqrt(36 + 64)

x = sqrt(100)

x = 10

Distance between tween my friends house and my house = 10 miles

Answer:

28.26

Step-by-step explanation:

6 divided by 2 = 3^2 = 9 x 3.14 = 28.26

The calculated value of the coefficient of determination is 0.073

From the question, we have the following parameters that can be used in our computation:

Regression = linear

Correlation coefficient, r, is -0.271

The coefficient of determination can be calculated using:

R = r²

Where

r = Correlation coefficient = -0.271

Substitute the known values in the above equation, so, we have the following representation

R = (-0.271)²

Evaluate the exponent

R = 0.073

Hence, the coefficient of determination is 0.073

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Answer:

Radius is the diameter divided by 2

Answer:

c = 1

Step-by-step explanation:

we have -1/6 + 7/6

since 6 is a common denominator we can do

[tex]\frac{-1}{6} +\frac{7}{6} = c\\\frac{-1+7}{6} = c\\\frac{6}{6} = c\\1 = c\\c = 1[/tex]

Answer:

B. 0

Step-by-step explanation:

There aren't enough sides for you to roll a nine

(a) The ordinary differential equation is given by y(t) + y(t) + 3y(t) = 0. Using Laplace transform, we have(L [y(t)] + L [y(t)] + 3L [y(t)]) = 0L [y(t)] (s + 1) + L [y(t)] (s + 1) + 3L [y(t)] = 0L [y(t)] (s + 1) = - 3L [y(t)]L [y(t)] = - 3L [y(t)] /(s + 1)Taking the inverse Laplace of both sides, we have y(t) = L -1 [- 3L [y(t)] /(s + 1)]y(t) = - 3L -1 [L [y(t)] /(s + 1)]

On comparison, we get y(t) = 3e^{-t} - 2e^{-3t}.The initial conditions are y(0) = 1 and y(0) = 2 respectively.(b) The ordinary differential equation is given by y(t) - 2y(t) + 4y(t) = 0. Using Laplace transform, we have L [y(t)] - 2L [y(t)] + 4L [y(t)] = 0L [y(t)] = 0/(s - 2) + (- 4)/(s - 2)

Taking the inverse Laplace of both sides, we have y(t) = L -1 [0/(s - 2) - 4/(s - 2)]y(t) = 4e^{2t}.The initial conditions are y(0) = 1 and y(0) = 2 respectively.(c) The ordinary differential equation is given by y(t) + y(t) = sint. Using Laplace transform, we have L [y(t)] + L [y(t)] = L [sint]L [y(t)] = L [sint]/(s + 1)

Taking the inverse Laplace of both sides, we have y(t) = L -1 [L [sint]/(s + 1)]y(t) = sin(t) - e^{-t}.The initial conditions are y(0) = 1 and y(0) = 2 respectively.(d) The ordinary differential equation is given by y(t) + 3y(t) = sint. Using Laplace transform, we have L [y(t)] + 3L [y(t)] = L [sint]L [y(t)] = L [sint]/(s + 3)Taking the inverse Laplace of both sides, we have y(t) = L -1 [L [sint]/(s + 3)]y(t) = (1/10)(sin(t) - 3cos(t)) - (1/10)e^{-3t}.

The initial conditions are y(0) = 1 and y(0) = 2 respectively.(e) The ordinary differential equation is given by y(t) + 2y(t) = e^{t}. Using Laplace transform, we have L [y(t)] + 2L [y(t)] = L [e^{t}]L [y(t)] = 1/(s + 2)Taking the inverse Laplace of both sides, we havey(t) = L -1 [1/(s + 2)]y(t) = e^{-2t}The initial conditions are y(0) = 1 and y(0) = 2 respectively.

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Answer:

d is parallel to e

Step-by-step explanation:

Since b is parallel to e and d is perpendicular to b , then

d is perpendicular to e

Two ordinary dice are thrown simultaneously. Determine the number of throws necessary to obtain at least once with probability 0.49 at least once the pair (6,6).

Solution: The probability of getting a pair of 6s in a single throw is 1/36.The probability of not getting a pair of 6s in a single throw is 1 - 1/36 = 35/36.

The probability of not getting a pair of 6s in n throws is (35/36)^n.

The probability of getting a pair of 6s in n throws is 1 - (35/36)^n.

So, for at least one pair of 6s with probability 0.49 in n throws, we have:

1 - (35/36)^n = 0.49⇒ (35/36)^n = 0.51⇒ n ln (35/36) = ln 0.51⇒ n = ln 0.51/ln (35/36) = 72.5 ~ 73So, at least 73 throws are necessary to obtain at least once with probability 0.49 at least once the pair (6,6).

Answer: At least 73 throws are necessary to obtain at least once with probability 0.49 at least once the pair (6,6).

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Answer:73 ft

Step-by-step explanation:

Answer:

(120 + 12.5pi) ft^2

Step-by-step explanation:

10ft x 12 ft = 120ft^2

10ft/2 = 5 ft (Radius)

Area of semi circle:

[tex]\frac{\pi r^{2} }{2} = \frac{\pi 5^{2} }{2} = 12.5\pi ft^{2}[/tex]

Area = (120 + 12.5pi) ft^2

The probability of a student passing the module is less than 0.75. Therefore, the lecturer should reconsider his method of teaching.

Question analysis A statistics module has been running for many years, and the module is given to n students each year. It has been discovered that in each year, the number of students passing the exam is distributed Bi(n, 0.75). In the current year, 150 students took the module for the first time, and 105 students passed the exam.

Using the 0.05 level of significance, we will conduct a hypothesis test to decide if the module's pass rate this year is less than 0.75.AssumptionsIf the number of trials is huge, the distribution of successes will be nearly normal. The number of trials n is greater than 30 in this situation. The probability of success in each trial is the same, namely p = 0.75. This condition is also satisfied. Therefore, we may use a normal distribution to approximate the binomial distribution.

What is the conclusion?

Null hypothesis: H₀: P = 0.75

Alternative hypothesis: H₁: P < 0.75The level of significance is 0.05, which implies that the rejection area will be in the left tail because the alternative hypothesis is one-tailed. Since the distribution of successes is approximately normal with a mean of np and a variance of np(1−p), we may find the p-value using this formula:
[The probability that X ≤ 105]
= [Z = (X − µ)/σ]
= [Z = (105 − (150 × 0.75))/sqrt(150 × 0.75 × (1 − 0.75))]
= [Z = (105 − 112.5)/3.2958]
= -2.2782
The p-value is [P(Z < -2.2782)] = 0.011. Because the p-value is less than 0.05, we reject the null hypothesis and accept the alternative hypothesis.

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Given : A statistics module has been running for many years and, in the past, it has been found that each year the number of students passing the exam has distribution Bi(n. 0.75), where n are the number of students taking the module that year. A lecturer is teaching the module for the first time and 105 out of 150 students pass the exam. The conclusion is that we fail to reject the null hypothesis.

The null and alternative hypotheses are given as follows:

Null hypothesis: The probability of a student passing the module is 0.75.

Alternative hypothesis: The probability of a student passing the module is less than 0.75.

We need to perform a hypothesis test at the 0.05-significance level.

The given probability distribution Bi(n,0.75) can be approximated to the normal distribution N(np,npq) under the following conditions:

The sample size n is large enough.

np≥5 and nq≥5, where q=1-p.

Here, n=150 and

p = 0.75

q = 1−p

= 1−0.75

= 0.25

Since np and nq are both greater than 5, the distribution Bi(150,0.75) can be approximated by the normal distribution N(150×0.75,150×0.75×0.25) = N(112.5,28.125).

Let X be the number of students that passed the module.

Under the null hypothesis, X follows the binomial distribution Bi(150,0.75).

Let μ be the mean of X under the null hypothesis.

μ = np

= 150×0.75

= 112.5

Since the alternative hypothesis is the probability of passing the module is less than 0.75, we need to perform a one-tailed test in the left tail at the 0.05-significance level.

The test statistic is given by,

Z=(X−μ)/σ

Z=(105−112.5)/√28.125/150

Z ≈ −1.5

This is a left-tailed test, so the critical value for a 0.05-significance level is z=−1.645.

Since the test statistic z=-1.5 > critical value z=-1.645, we fail to reject the null hypothesis.

Hence, there is not enough statistical evidence to conclude that the probability of a student passing the module is less than 0.75.

Therefore, the conclusion is that we fail to reject the null hypothesis.

To know more about probability distribution, visit:

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