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Stells [14]
3 years ago
6

A container ship has to make 2 deliveries then return to the pier it started from. The boat travels 8 miles straight north to it

s first stop. The nexi delivery is directly across the river 3 miles straight west. What angle must the boat travel at to get back to the pier it started from?​
Mathematics
1 answer:
oksano4ka [1.4K]3 years ago
6 0
Straight north is the answer
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I don't know if you still need help. but the answer is the first one.
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3 years ago
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Evaluate i^31 please help
Nadya [2.5K]

ANSWER

{i}^{31}  =  - i

EXPLANATION

We want to evaluate

{i}^{31}

Use indices to rewrite the expression as:

=  {i}^{30}  \times i

We know that

{i}^{2}  =  - 1

So we rewrite the expression to obtain;

=  ({ {i}^{2}) }^{15}  \times i

This gives us;

=   {( - 1) }^{15}  \times i

This simplifies to

=  - 1 \times i

=  - i

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3 years ago
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It’s B correct answer
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2 years ago
A government report gives a 99% confidence interval for the proportion of welfare recipients who have been receiving welfare ben
juin [17]

Answer:

The estimation for the proportion for this case is \hat p = 0.21

And we know that the 99% confidence interval is given by:

0.21 \pm 0.045

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

The confidence interval would be given by this formula  

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=1.96

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.58

The estimation for the proportion for this case is \hat p = 0.21

And we know that the 99% confidence interval is given by:

0.21 \pm 0.045

So then the margin of error at 99% of confidence is 0.045, and the margin of error is given by this formula:

ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}  

If we decrease the confidence level this margin of error can't be higher than 0.045 so then the correct answer for this case would be:

d. 21% ± 4.8%

Because if the z value decrease from 2.58 to 1.96 not makes sense that the margin of error increase from 0.045(4.5%) to 0.048(4.8%)

5 0
3 years ago
CAN YALL ANSWER MY BROTHERS HW BEC HIS DEADAZZ DOENST WANT TO BTW HE IS IN 3RD GRADE
nadezda [96]
You should try to retake the picture can’t c anything
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2 years ago
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