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Ad libitum [116K]
3 years ago
10

You are applying fertilizer to a football field. The field is 360 feet long and 160 feet wide. You use 8 pounds of fertilizer pe

r 1,000 square feet. The fertilizer comes in 50-pound bags. How many bags of fertilizer will you need to complete the job?
Mathematics
1 answer:
Assoli18 [71]3 years ago
6 0

Answer: 10 bags

Step-by-step explanation:

First we must calculate the field area. Assuming it is a rectangle then its area equals the product of its length times its width

A =360*160\\\\A = 57600 ft^2

You use 8 pounds of fertilizer per 1,000 square feet

So the amount of fertilizer you should use is:

57,600 ft^2 * \frac{8\ pounds}{1,000\ ft^2} = 460.8\ pounds

As the fertilizer comes in bags of 50 pound then you need

460.8\ pounds * \frac{1\ bag}{50\ pounds} = 9.2\ bags

Since you can not buy 9.2 bags then we round up to the next whole number.

Finally you need 10 bags of fertilizer

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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
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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.

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