How many 3 letter permutations are possible for the letters s t u d y h a r d

Can somebody help me with the rest of the question

kari74 [83]

Shop #1 has the best price for 21 pizzas because it is $105, while Shop #2 costs $147 for 21 pizzas which is a lot more money. they would save $42, which is 147-105=42. hope this helps:)

5 0

3 years ago

State the range of the relation {(1, -1), (2, 3), (3, 5), (4, 8)}.

DiKsa [7]

The range would be -1 to 8

8 0

4 years ago

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 30,393 miles, with a standard

Pie

Answer:

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem:

Normal probability distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 30393, \sigma = 2876, n = 37, s = \frac{2876}{\sqrt{37}} = 472.81$

What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

This probability is the pvalue of Z when X = 30393 + 339 = 30732 subtracted by the pvalue of Z when X = 30393 - 339 = 30054. So

X = 30732

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

By the Central Limit Theorem

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

$Z = \frac{30732 - 30393}{472.81}$

$Z = 0.72$

$Z = 0.72$ has a pvalue of 0.7642.

X = 30054

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

$Z = \frac{30054 - 30393}{472.81}$

$Z = -0.72$

$Z = -0.72$ has a pvalue of 0.2358

0.7642 - 0.2358 = 0.5284

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

4 0

4 years ago

A Ski resort tracks the proportion of seasonal employees who are rehired each season. Rehiring a seasonal employee is beneficial

DaniilM [7]

The test statistic z will be equal to -0.946 and it shows that there is no significant difference in the proportion of rehires between full time and part time.

Given sample sizes of 833 and 386 and result of samples 434 and 189.

$n_{1} =833, n_{2}=386$

Proportion of full time=434/833=0.52

Proportion of part time=189/386=0.49.

Difference in proportion =0.52-0.49

$H_{0}:$ TTF- i∈ rho=0

$H_{1}:$ TTF+i∈ rho≠0.

Mean of difference=0.03

Z=(X-μ)/σ

σ= $\sqrt{0.4517/447}$

=0.0317

σ=0.0317

z=(0-0.03)/0.0317

=-0.03/0.0317

=-0.317

p value will be =0.1736.

Because p value is greater than 0.01 so we will accept the null hypothesis which shows that there is no significant difference in the proportions.

Hence there is no significant difference in the proportion of rehires between full time and part time.

Learn more about z test at brainly.com/question/14453510

#SPJ4

6 0

2 years ago

How to factor 5y+11y+2

aksik [14]

16y+2 is your answer

7 0

3 years ago