A cake recipe calls for 2 and 5/6 cups of sugar. A caterer has 13 1/2

What is the probability of choosing a burrito with cheese and a taco or wrap with salsa??

torisob [31]

Step-by-step explanation:

50/50 chance of choosing one or the other because there is only two options

6 0

2 years ago

A construction company is considering submitting bids for contracts of three different projects. The company estimates that it h

julsineya [31]

Answer:

a. $P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

b. $E(x) = 0.3$

c. $S(x)=0.5196$

d. $E=5,000$

Step-by-step explanation:

The probability that the company won x bids follows a binomial distribution because we have n identical and independent experiments with a probability p of success and (1-p) of fail.

So, the PMF of X is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

Where p is 0.1 and it is the chance of winning. Additionally, n is 3 and it is the number of bids. So the PMF of X is:

$P(x)=\frac{3!}{x!(3-x)!}*0.1^{x}*(1-0.1)^{n-x}\\$

For binomial distribution:

$E(x)=np\\S(x)=\sqrt{np(1-p)}$

Therefore, the company can expect to win 0.3 bids and it is calculated as:

$E(x) = np = 3*0.1 = 0.3$

Additionally, the standard deviation of the number of bids won is:

$S(x)=\sqrt{np(1-p)}=\sqrt{3(0.1)(1-0.1)}=0.5196$

Finally, the probability to won 1, 2 or 3 bids is equal to:

$P(1)=\frac{3!}{1!(3-1)!}*0.1^{1}*(1-0.1)^{3-1}=0.243\\P(2)=\frac{3!}{2!(3-2)!}*0.1^{2}*(1-0.1)^{3-2}=0.027\\P(3)=\frac{3!}{3!(3-3)!}*0.1^{3}*(1-0.1)^{3-3}=0.001$

So, the expected profit for the company is equal to:

$E=-10,000+50,000(0.243)+100,000(0.027)+150,000(0.001)\\E=5,000$

Because there is a probability of 0.243 to win one bid and it will produce 50,000 of income, there is a probability of 0.027 to win 2 bids and it will produce 100,000 of income and there is a probability of 0.001 to win 3 bids and it will produce 150,000 of income.

5 0

3 years ago

Given WXYZ what is the measure of Z

nekit [7.7K]

Answer: 35

Step-by-step explanation:

5 0

2 years ago

The manager of the BiLo Supermarket in Mt. Pleasant, Rhode Island, gathered the following information on the number of times a c

pishuonlain [190]

Answer: kindly check explanation

Step-by-step explanation:

Given the data :

5 3 3 1 4 4 5 6 4 2 6 6 6 7 1 1 14 1 2 4 4 4 5 6 3 5 3 4 5 6 8 4 7 6 5 9 11 3 12 4 7 6 5 15 1 1 10 8 9 2 12

Frequency and relative frequency table can be found in the attachment below.

b. Where do the data tend to cluster?

3 up to 6

7 0

3 years ago

There are 20 machines in a factory. 7 of the machines are defective.

adelina 88 [10]

Answer:

0.1225

Step-by-step explanation:

Given

Number of Machines = 20

Defective Machines = 7

Required

Probability that two selected (with replacement) are defective.

The first step is to define an event that a machine will be defective.

Let M represent the selected machine sis defective.

P(M) = 7/20

Provided that the two selected machines are replaced;

The probability is calculated as thus

P(Both) = P(First Defect) * P(Second Defect)

From tge question, we understand that each selection is replaced before another selection is made.

This means that the probability of first selection and the probability of second selection are independent.

And as such;

P(First Defect) = P (Second Defect) = P(M) = 7/20

So;

P(Both) = P(First Defect) * P(Second Defect)

PBoth) = 7/20 * 7/20

P(Both) = 49/400

P(Both) = 0.1225

Hence, the probability that both choices will be defective machines is 0.1225

4 0

3 years ago