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Katyanochek1 [597]

3 years ago

15

DOES ANYONE KNOW HOW MANY QUESTION ARE ON THE CUMULATIVE EXAM ON EDGENUITY????? AND IM GONNA NEED LOTS OF HELP. ALSO WILL MARK B

RAINLIEST!!

1 answer:

Sauron [17]3 years ago

4 0

Answer:

i think 60!!!!!!!!! i took it and i think it was sixty questions!!!!! good luck

Step-by-step explanation:

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3 years ago

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Finding the value of the x

natali 33 [55]

Answer: do a proportion

Step-by-step explanation:

So it would be a number over a number and equals X over a number.

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3 years ago

What is the slope of a line that is perpendicular to y+8=−32(x−2)

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boonpono

Step-by-step explanation:

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3 years ago

2. The National Safety Council routinely analyzes the benefit of seat belt use on driver safety. Their data showed that among 28

JulijaS [17]

Answer:

We conclude that there is difference in the proportion of deaths between the 2 groups.

Step-by-step explanation:

We are given that among 2823 drivers not wearing seat belts, 31 died as a result of injuries, and among 7765 drivers wearing seat belts 16 were killed.

Let $p_1$ = <u><em>proportion of deaths when drivers were not wearing seat belts.</em></u>

$p_2$ = <u><em>proportion of deaths when drivers were wearing seat belts.</em></u>

So, Null Hypothesis, $H_0$ : $p_1=p_2$ {means that there is no difference in the proportion of deaths between the 2 groups}

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$ {means that there is difference in the proportion of deaths between the 2 groups}

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 deaths when drivers were not wearing seat belts = $\frac{31}{2823}$ = 0.011

$\hat p_2$ = sample proportion of deaths when drivers were wearing seat belts = $\frac{16}{7765}$ = 0.002

$n_1$ = sample of drivers not wearing seat belts = 2823

$n_2$ = sample of drivers wearing seat belts = 7765

So, <u><em>the test statistics</em></u> = $\frac{(0.011-0.002)-(0)}{\sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} } }$

= 4.438

The value of z test statistics is 4.438.

<u>Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</u>

Since our test statistic doesn't lie within the range of critical values of z, 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 there is difference in the proportion of deaths between the 2 groups.

Also, <u>Margin of error</u> (E) = $1.96 \times \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }$

= $1.96 \times \sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} }$

= <u>0.00397</u>

5 0

3 years ago

Find the percent of increase or decrease?<br> N to 4/5n?

d1i1m1o1n [39]

Answer:

Decrease by 20%.

Step-by-step explanation:

We have to find the percent of increase or decrease from n to $\frac{4}{5}n$ .

Now, $\frac{4}{5}n = 0.8n < n$

So there is a decrease fro n to $\frac{4}{5}n$ .

Now, the percentage of decrease will be given by

= $[ \frac{\textrm {Original Value - Final Value}}{\textrm {Original Value}}]\times 100 \%$

= $[ \frac{n - 0.8n}{n}]\times 100 \%$

= 20% (Answer)

8 0

3 years ago