A 35-foot tall pole is leading towards a building. To stop the pole from leaning

Mathematics

1 answer:

Lapatulllka [165]2 years ago

5 0

9514 1404 393

Answer:

38.5°

Step-by-step explanation:

You are given all three side lengths of the relevant triangle, so the Law of Cosines can be used to find any desired angle. If we call the desired angle C, then that law tells you ...

C = arccos((a² +b² -c²)/(2ab))

where a and b are the triangle side lengths adjacent to the angle of interest, and c is the side opposite. Here, 'a' and 'b' are the pole and cable lengths, and 'c' is the distance of the stake from the pole.

C = arccos((35² +48² -30²)/(2·35·48)) = arccos(2629/3360)

C ≈ 38.515°

The cable makes an angle of about 38.5° with the pole.

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There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

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".

The population proportion have the following distribution

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

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

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 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.