Choose the equation below that represents the line passing though the point(-5, 1) with a slope of 2/3

Daniel's tennis team made $582.85 from running the concession stand at a basketball game. The food and drinks cost the team $64.

Flura [38]

Let x be the number of members in Daniel's tennis team.

The amount made = $582.85

Expenses (food and drinks) = $64

Balance remaining for sharing = 582.85 - 64 = $518.85

Amount received by each player = (amount made - Expenses) / Number of members = 518.85/x

8 0

3 years ago

Read 2 more answers

Which container holds more, a Elgin millimeter carton, or a 1 L bottle

anyanavicka [17]

The answer would be a 1 Liter bottle, because a liter is 33 oz rounded and and Elgin millimeter container is 25 oz.

7 0

3 years ago

Find the 94th term of the arithmetic sequence 25, 31, 37...

avanturin [10]

Answer:

589 is da answer

Step-by-step explanation:

25+6(94)=?

6 0

2 years ago

uppose that Mary's utility function is U(W) = W0.5, where W is wealth. She has an initial wealth of $100. How much of a risk

Stels [109]

Note that $U(W) = W^{0.5}$

Answer:

Mary's risk premium is $0.9375

Step-by-step explanation:

Mary's utility function, $U(W) = W^{0.5}$

Mary's initial wealth = $100

The gamble has a 50% probability of raising her wealth to $115 and a 50% probability of lowering it to $77

Expected wealth of Mary, $E_w$

$E_{w}$ = (0.5 * $115) + (0.5 * $77)

$E_{w}$ = 57.5 + 38.5

$E_{w}$ = $96

The expected value of Mary's wealth is $96

Calculate the expected utility (EU) of Mary:-

$E_u = [0.5 * U(115)] + [0.5 * U(77)]\\E_u = [0.5 * 115^{0.5}] + [0.5 * 77^{0.5}]\\E_u = 5.36 + 4.39\\E_u = \$ 9.75$

The expected utility of Mary is $9.75

Mary will be willing to pay an amount P as risk premium to avoid taking the risk, where

U(EW - P) is equal to Mary's expected utility from the risky gamble.

U(EW - P) = EU

U(94 - P) = 9.63

Square root (94 - P) = 9.63

If Mary's risk premium is P, the expected utility will be given by the formula:

$E_{u} = U(E_{w} - P)\\E_{u} = U(96 - P)\\E_u = (96 - P)^{0.5}\\(E_u)^2 = 96 - P\\ 9.75^2 = 96 - P\\95.0625 = 96 - P\\P = 96 - 95.0625\\P = 0.9375$

Mary's risk premium is $0.9375

7 0

2 years ago

When grading an exam, 90% of a professor's 50 students passed. If the professor randomly selected 10 exams, what is the probabil

Damm [24]

Using the binomial distribution, it is found that there is a:

a) 0.9298 = 92.98% probability that at least 8 of them passed.

b) 0.0001 = 0.01% probability that fewer than 5 passed.

For each student, there are only two possible outcomes, either they passed, or they did not pass. The probability of a student passing is independent of any other student, hence, the binomial distribution is used to solve this question.

<h3>What is the binomial probability distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem:

90% of the students passed, hence $p = 0.9$ .

The professor randomly selected 10 exams, hence $n = 10$ .

Item a:

The probability is:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{10,8}.(0.9)^{8}.(0.1)^{2} = 0.1937$

$P(X = 9) = C_{10,9}.(0.9)^{9}.(0.1)^{1} = 0.3874$

$P(X = 10) = C_{10,10}.(0.9)^{10}.(0.1)^{0} = 0.3487$

Then:

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1937 + 0.3874 + 0.3487 = 0.9298$

0.9298 = 92.98% probability that at least 8 of them passed.

Item b:

The probability is:

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Using the binomial formula, as in item a, to find each probability, then adding them, it is found that:

$P(X < 5) = 0.0001$

Hence:

0.0001 = 0.01% probability that fewer than 5 passed.

You can learn more about the the binomial distribution at brainly.com/question/24863377

3 0

1 year ago