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3 years ago

1) 8n - 40 = 6(8 + 5n)

2 answers:

8 0

$\text{Hello there!}\\\\\text{Solve for n:}\\\\8n - 40 = 6(8 + 5n)\\\\\text{First, distribute the 6 to the variables inside the parenthesis}\\\\8n-40=48+30n\\\\\text{Subtract both sides by -30n}\\\\-22n-40=48\\\\\text{Add 48 to both sides}\\\\-22n=88\\\\\text{Divide both sides by -22}\\\\n=-4\\\\\boxed{n=-4}$

xxMikexx [17]3 years ago

7 0

Answer:

Step-by-step explanation:

Isolate n on one side of the equation.

DISTRIBUTIVE PROPERTY:

A(B+C)=AB+AC

A(B-C)=AB-AC

6(8+5n) (First, expand by using distributive property.)

6*8=48

6*5n=30n

48+30n

8n-40=48+30n

8n-40+40=48+30n+40 (Next, add 40 from both sides.)

48+40 (Solve.)

48+40=88

8n=30n+88

8n-30n=30n+88-30n (Subtract 30 from both sides.)

8n-30n (Solve.)

8n-30n=-22n

-22n=88

-22n/-22=88/-22 (Then, divide by -22 from both sides.)

88/-22 (Solve.)

88/-22=-4

n=-4

Therefore, the final answer is n=-4.

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Step-by-step explanation:

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3 years ago

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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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 $p_1$ = percentage of employed workers who have registered to vote.

$p_2$ = percentage of unemployed workers who have registered to vote.

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 Two-sample z test for proportions;

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, the test statistics = $\frac{(0.56-0.46)-(0)}{\sqrt{\frac{0.56(1-0.56)}{513}+\frac{0.46(1-0.46)}{604} } }$

= 3.349

The value of z test statistics is 3.349.

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

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 we reject our null hypothesis.

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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