What is the surface area of the prism?

The University of Washington claims that it graduates 85% of its basketball players. An NCAA investigation about the graduation

Nonamiya [84]

Probabilities are used to determine the chances of events

The given parameters are:

Sample size: n = 20
Proportion: p = 85%

<h3>(a) What is the probability that 11 out of the 20 would graduate? </h3>

Using the binomial probability formula, we have:

$P(X = x) = ^nC_x p^x(1 - p)^{n -x}$

So, the equation becomes

$P(x = 11) = ^{20}C_{11} \times (85\%)^{11} \times (1 - 85\%)^{20 -11}$

This gives

$P(x = 11) = 167960 \times (0.85)^{11} \times 0.15^{9}$

$P(x = 11) = 0.0011$

Express as percentage

$P(x = 11) = 0.11\%$

Hence, the probability that 11 out of the 20 would graduate is 0.11%

<h3>(b) To what extent do you think the university’s claim is true?</h3>

The probability 0.11% is less than 50%.

Hence, the extent that the university’s claim is true is very low

<h3>(c) What is the probability that all 20 would graduate? </h3>

Using the binomial probability formula, we have:

$P(X = x) = ^nC_x p^x(1 - p)^{n -x}$

So, the equation becomes

$P(x = 20) = ^{20}C_{20} \times (85\%)^{20} \times (1 - 85\%)^{20 -20}$

This gives

$P(x = 20) = 1 \times (0.85)^{20} \times (0.15\%)^0$

$P(x = 20) = 0.0388$

Express as percentage

$P(x = 20) = 3.88\%$

Hence, the probability that all 20 would graduate is 3.88%

<h3>(d) The mean and the standard deviation</h3>

The mean is calculated as:

$\mu = np$

So, we have:

$\mu = 20 \times 85\%$

$\mu = 17$

The standard deviation is calculated as:

$\sigma = np(1 - p)$

So, we have:

$\sigma = 20 \times 85\% \times (1 - 85\%)$

$\sigma = 20 \times 0.85 \times 0.15$

$\sigma = 2.55$

Hence, the mean and the standard deviation are 17 and 2.55, respectively.

Read more about probabilities at:

brainly.com/question/15246027

8 0

3 years ago

What is 765903 rounded to the nearest thousand

Dominik [7]

765,000 should be the correct answer

3 0

4 years ago

Read 2 more answers

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is r

svp [43]

Answer:

0.1 = 10% probability that the class length is between 51.5 and 51.7 min, that is, P(51.5 < X < 51.7) = 0.1.

Step-by-step explanation:

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distributon has two bounds, a and b, and the probability of finding a value between c and d is given by:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min.

This means that $a = 50, b = 52$

If one such class is randomly selected, find the probability that the class length is between 51.5 and 51.7 min.

$P(51.5 \leq X \leq 51.7) = \frac{51.7 - 51.5}{52 - 50} = \frac{0.2}{2} = 0.1$

0.1 = 10% probability that the class length is between 51.5 and 51.7 min, that is, P(51.5 < X < 51.7) = 0.1.

3 0

3 years ago

I need help please and thank you

Brums [2.3K]

Well I THINK triangle ABC is just a stretched out version of triangle DEF.

To find the area of a triangle, you multiply base and height, and divide by 2.

Area of DEF is 6, so that means that (4 * x)/2 = 6.

6*2 = 12....

4 * x = 12........

4 * 3 = 12.

x = 3.

Comparing DEF to ABC: ABC's base is 3 times as long as DEF's. So that means the height must be three times as long.

3*3 = 9.

(12 * 9)/2 =54

5 0

3 years ago

Read 2 more answers

The cost to rent a surfboard at the beach is $10.25 an hour plus an insurance fee of $25. Keith spent $55.75 when renting a surf

goblinko [34]

Answer: Keith spent 3 hours renting the surfboard.

Step-by-step explanation:

h=hour

10.25h+25=55.75

55.75-25= 30.75

30.75/10.25= 3

and if you want to check you can do

10.25(3)+25=

which would equal to 55.75

4 0

2 years ago