Clive and Erica both have part-time jobs where they get paid by the hour. Erica makes more than Clive. The number of dollars, dd, Clive makes in hh hours can be modeled by the equation d=9.5hd=9.5h. Erica created a graph, with the number of hours she works along the xx-axis and the number of dollars she makes along the yy-axis. Her graph is a line with a slope of 99 and a yy-intercept is 00. Which is a true statement?

Missing Options:

A. Erica's graph is incorrect, because the slope  should be greater than 9.5.

B. Erica's graph is incorrect, because the y-intercept should be greater than 9.5.

C. Erica's graph is correct, because the slope  should be less than 9.5.

D. Erica's graph is correct, because the y–intercept  should be less than 9.5.

A. Erica's graph is incorrect, because the slope  should be greater than 9.5.

Step-by-step explanation:

Given

Clive:

Erica:

--- slope

--- y intercept

First, we need to determine the equation of Erica's earnings using

Where

m = slope = 9

b = y intercept = 0

So, the equation becomes

This implies that Erica earns 9 dollars per hour

For Clive:

This means that Clive earns 9.5 dollars per hour

From the question, we understand that Erica earns more than Clive.

So, we can conclude that Erica's graph is incorrect because the slope must be greater than 9.5.

When the slope is greater than 9.5, then she earns  more than Clive.

We reject H₀

Step-by-step explanation:

Normal Distribution with  n  = 22   ( sample size)

n<30 we need to use a t-student distribution

Sample mean  μ =   396        

Population mean  ( required mean )  μ₀  = 405

Sample variance is  441 then sample standard deviation s = √441

s  =  21

Hypothesis Test

Null Hypothesis                                          H₀         μ   =   μ₀

Alternative Hypothesis                               Hₐ       μ   <   μ₀

Significance level   α = 0,1    

The test is a one-tail test to the left

From t-student table we find for  t(c)

degree of fredom     df  = 22 -1       df  =  21

and  α = 0,1           t(c)  =  - 1,3232

To compute t(s)

t(s)  =  ( μ  -  μ₀ ) / s /√n

t(s)  =  ( 396  - 405 )*√n / 21

t(s)  =  - 9 *4,69 /21

t(s)  =  -  2,01

Comparing

t(s) and  t(c)          - 2,01     and   - 1,3232

|t(s)| > |t(c)|   then   t(s) is in the rejection region we must reject H₀