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Alexxandr [17]
3 years ago
11

Sarah has been running a dog-walking business since 2010. She walks dogs twice a day, takes them to the park, and returns them t

o their homes. Each year, she has increased her fee by the same amount. The table shows what Sarah charged each customer for two given years of her business:
Year Annual Dog-walking Fee
2010 $350
2014 $750


A. What is the rate of change and initial value for Sarah’s business? How do you know?
B. Write an equation in slope-intercept form to represent the fees that Sarah charges each year.
Mathematics
1 answer:
Contact [7]3 years ago
4 0
For A, you wanna look at $350 and $750 and find out how much her pay increased from 2010 to 2014.
For B, you want to create an equation (math problem) <span>to represent the fees that Sarah charges each year that would be reliant to your answer for part A. Hope this helps!
</span><span>
</span>
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The U.S. Census Bureau conducts annual surveys to obtain information on the percentage of the voting-age population that is regi
yan [13]

Answer:

We conclude that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote.

Step-by-step explanation:

We are given that 513 employed persons and 604 unemployed persons are independently and randomly selected, and that 287 of the employed persons and 280 of the unemployed persons have registered to vote.

Let p_1 = <u><em>percentage of employed workers who have registered to vote.</em></u>

p_2 = <u><em>percentage of unemployed workers who have registered to vote.</em></u>

So, Null Hypothesis, H_0 : p_1\leq p_2      {means that the percentage of employed workers who have registered to vote does not exceeds the percentage of unemployed workers who have registered to vote}

Alternate Hypothesis, H_A : p_1>p_2     {means that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote}

The test statistics that would be used here <u>Two-sample z test for proportions;</u>

                          T.S. =  \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }  ~ N(0,1)

where, \hat p_1 = sample proportion of employed workers who have registered to vote = \frac{287}{513} = 0.56

\hat p_2 = sample proportion of unemployed workers who have registered to vote = \frac{280}{604} = 0.46

n_1 = sample of employed persons = 513

n_2 = sample of unemployed persons = 604

So, <u><em>the test statistics</em></u>  =  \frac{(0.56-0.46)-(0)}{\sqrt{\frac{0.56(1-0.56)}{513}+\frac{0.46(1-0.46)}{604} } }

                                       =  3.349

The value of z test statistics is 3.349.

<u>Now, at 0.05 significance level the z table gives critical value of 1.645 for right-tailed test.</u>

Since our test statistic is more than the critical value of z as 3.349 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the percentage of employed workers who have registered to vote exceeds the percentage of unemployed workers who have registered to vote.

5 0
3 years ago
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