Pleaaase helppp meee im being timedd please explain too and i will be marking brainliest thank u :)

Sterre charges $35 to file tax returns, but files for free if she only needs the easiest form. Then she donates $2 to clean wate

Sauron [17]

Charges for filing each tax return not through the easiest form = $35

Charges for the filing each tax return through the easiest form = $0

Donation made to clean water project for each tax return filed = $2

Let total tax returns filed be 'a' and total tax returns filed at $35 be 'b'

⇒ Tax returns filed at $0 = a-b

Total money charged by Sterre = $7, 245

Total donation made by Sterre = $1, 242

Now, first we will determine the total tax returns filed by Sterre

Total donation made by Sterre = Donation made per tax return × Total returns filed by Sterre

⇒ 1,242 = 2 × a

⇒ a = 621

Hence, Sterre filed total of 621 tax returns

Now total money charged by Sterre = Charges for filing each tax return not through the easiest form × Tax returns filed at $35

⇒ 7,245 = 35 * b

⇒ b = 207

Hence, Sterre filed 207 returns at $35.

⇒ Tax returns filed at $0 = 621 - 207

⇒ Tax returns filed at $0 = 414

Hence, at $35, Sterre filed 207 tax returns, and at $0 Sterre filed 414 tax returns.

6 0

4 years ago

What is 10× greater than 760

Anna007 [38]

Just ad a zero to the end and you get 7600

3 0

3 years ago

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Two numbers have sum of 100 and a difference of 6

Hunter-Best [27]

The answer to your question is 53 and 47

6 0

3 years ago

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I don't get it please help

Elis [28]

2, (double) 4, (double) 8 (double) 16, So 16 in week 4
2+4+8+16 = 30

5 0

3 years ago

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Ignoring those who said they weren't sure, there were 297 men asked, and 183 said yes, they had driven a car when they probably

USPshnik [31]

Answer:

$z=\frac{0.616-0.5}{\sqrt{\frac{0.5(1-0.5)}{297}}}=3.998$

$p_v =2*P(z>3.998)=0.0000639$

With the most common significance levels used $\alpha= 0.1, 0.05, 0.01$ we see that the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can say that the true proportion is significantly higher than 0.5

Step-by-step explanation:

Information given

n=297 represent the random sample of male taken

X=183 represent the men who said yes, they had driven a car when they probably had too much alcohol

$\hat p=\frac{183}{297}=0.616$ estimated proportion of men who said yes, they had driven a car when they probably had too much alcohol

$p_o=0.5$ is the value that we want to test

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Hypothesis to test

We need to conduct a hypothesis in order to test the claim that the majority of men in the population (that is, more than half) would say that they had driven a car when they probably had too much alcohol, and the system of hypothesis are:

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: $p > 0.5$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

After replace we got:

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.616-0.5}{\sqrt{\frac{0.5(1-0.5)}{297}}}=3.998$

Decision

We have a right tailed test so then the p value would be:

$p_v =2*P(z>3.998)=0.0000639$

With the most common significance levels used $\alpha= 0.1, 0.05, 0.01$ we see that the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can say that the true proportion is significantly higher than 0.5

7 0

3 years ago