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3 years ago

PLEASE HELP WITH QUESTION! MARKING BRAINLIEST + POINTS GIVEN.

1 answer:

8 0

The simple answer (not using multiples of 2 $\pi$ (360°)) is
$= \sqrt[3]{125} ((cos( \frac{288}{3})) + isin(( \frac{288}{3})))
= 5(cos96 + isin96)$

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Lin took 75 boxes of carrots and 80 boxes of potatoes to the market. She sold 68 boxes of carrots and 71 boxes of potatoes. How

Lady bird [3.3K]

Answer:

Step-by-step explanation:

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4 years ago

Read 2 more answers

There are 9,481 eligible voters in a precinct. 500 were selected at random and asked to indicate whether they planned to vote fo

maria [59]

Answer:

The confidence limits for the proportion that plan to vote for the Democratic incumbent are 0.725 and 0.775.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of the 500 surveyed, 350 said they were going to vote for the Democratic incumbent.

This means that $n = 500, \pi = \frac{350}{500} = 0.75$

80% confidence level

So $\alpha = 0.2$ , z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$ , so $Z = 1.28$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.75 - 1.28\sqrt{\frac{0.75*0.25}{500}} = 0.725$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.75 + 1.28\sqrt{\frac{0.75*0.25}{500}} = 0.775$

The confidence limits for the proportion that plan to vote for the Democratic incumbent are 0.725 and 0.775.

8 0

3 years ago

I'm sooo confused...................................

user100 [1]

Answer:

8.14 is the answer

Step-by-step explanation:

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3 years ago

A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the

EastWind [94]

Answer:

a) $\hat p=\frac{471}{1024}=0.460$

The standard error is given by:

$SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156$

And the margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305$

b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

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

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

Data given and notation

n=1024 represent the random sample taken

X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

$\hat p=\frac{471}{1024}=0.460$ estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

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.