How long does it take for 1/2 cup of water to freeze?

Mahmoud earns $450 per week plus a 20% commission as a car salesman. He wants his

frosja888 [35]

The complete question implies that we define an inequality that represents the relationship between the number of hours worked in a week and the weekly sales

The inequality that relates the number of hours to the weekly sales is:

$450 + 0.20x \ge 35y$

We make use of the following representation:

$x \to$ weekly sales from cars.

$y \to$ hours worked in a week

His weekly salary is then calculated as:

Salary (S) = Earnings per week + Commission * Sales from car

So, we have:

$S = 450 + 20\% * x$

Express percentage as decimal

$S = 450 + 0.20* x$

$S = 450 + 0.20x$

Assume he works for y hours in a week.

His hourly rate is:

$Hourly = \frac{S}{y}$ --- i.e. weekly salary divided by number of hours

$Hourly = \frac{450 + 0.20x}{y}$

For this rate to be at least $\$35$ , the following condition must be true

$Hourly \ge 35$ --- i.e. is hourly rate must be greater than or equal 35

So, we have:

$\frac{450 + 0.20x}{y} \ge 35$

Multiply both sides by y

$450 + 0.20x \ge 35y$

5 0

3 years ago

NEED HELP WITH THIS PLSSS!!!!

rewona [7]

Answer:

use the formula with the width and height number

5 0

2 years ago

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be 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.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So $\mu = 373, \sigma = 67$

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So $n = 5, s = \frac{67}{\sqrt{5}} = 29.96$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 373}{29.96}$

$Z = 1.23$

$Z = 1.23$ has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So $\mu = 526, \sigma = 107$

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So $n = 5, s = \frac{107}{\sqrt{5}} = 47.86$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 526}{47.86}$

$Z = -2.42$

$Z = -2.42$ has a pvalue of 0.0078.

So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

7 0

3 years ago

Put thes fallowing numbers in order from largest to smallest -3, -3.5, 6, -6, 0, 1

Verdich [7]

6, 1, 0, -3, -3.5, -6

5 0

3 years ago

Read 2 more answers

A survey found that women's heights are normally distributed with mean 62.8 inches and standard deviation of 2.5 inches. This su

Jet001 [13]

<span>a)

Z*_Upper = (76 - 62.7)/2.5 = 5.32

Z*_Lower = (57 - 62.7)/2.5 = -2.28

The requirement is to get p(-2.28 < Z < 5.32) = p(Z<5.32) - p(Z<-2.28).

Use normal distribution table to get the answer for p and multiply with 100 to get the percentage.

The other questions are now easy for you to answer on your own. Hope it helps.

</span>

4 0

2 years ago

Read 2 more answers