Help can someone check

A particular employee arrives at work sometime between 8:00 a.m. and 8:30 a.m. Based on past experience the company has determin

Gala2k [10]

Answer:

0.3333 = 33.33% probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.

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:

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

A particular employee arrives at work sometime between 8:00 a.m. and 8:30 a.m.

We can consider 8 am = 0, and 8:30 am = 30, so $a = 0, b = 30$

Find the probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.

Between 15 and 25, so:

$P(15 \leq X \leq 25) = \frac{25 - 15}{30 - 0} = 0.3333$

0.3333 = 33.33% probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.

8 0

3 years ago

Select the correct numerator 7/9 - ?/27

damaskus [11]

Answer:

21

Step-by-step explanation:

27÷3 is 9 and 21÷3 is 7

7 0

3 years ago

Please help thank you

Eduardwww [97]

Answer:

its (C) or (B) but my answer would be (B)

Step-by-step explanation:

3 0

3 years ago

There are 1265 students at Cypress Middle School. 42% of the students are enrolled in PE. Which equation could be used to repres

Dovator [93]

Answer:

0.42* 1265=x

Step-by-step explanation:

There are 1265 students in Cypress Middle School. So we already know one part of the equation. 42% of these students are in PE. So we know that 42% of these 1265 students are in PE. Of these means you probably have to multiply 42% and 1265. 42% = 0.42. So to write an equation we take the 0.42 and 1265 and put it together. Therefore, 0.42 times 1265=x.

7 0

3 years ago

Suppose you can somehow choose two people at random who took the SAT in 2014. A reminder that scores were Normally distributed w

Sindrei [870]

Answer:

22.29% probability that both of them scored above a 1520

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 1497, \sigma = 322$

The first step to solve the question is find the probability that a student has of scoring above 1520, which is 1 subtracted by the pvalue of Z when X = 1520.

So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1520 - 1497}{322}$

$Z = 0.07$

$Z = 0.07$ has a pvalue of 0.5279

1 - 0.5279 = 0.4721

Each students has a 0.4721 probability of scoring above 1520.

What is the probability that both of them scored above a 1520?

Each students has a 0.4721 probability of scoring above 1520. So

$P = 0.4721*0.4721 = 0.2229$

22.29% probability that both of them scored above a 1520

8 0

3 years ago