Write in terms of i. Simplify as much as possible. -√-45

Mathematics

1 answer:

marusya05 [52]3 years ago

4 0

Answer:

I got - 6.70820393 i

Step-by-step explanation:

let me know if I am wrong :)

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How do it do this ????

V125BC [204]

Because he drove 1542.75 miles in 3 days, you would have to divide 1542.75 by 3 to figure out how many miles he drove in one day.

1542.75/3 = 514.25

Then to figure out how many miles he drove per hour, you would divide 514.25 by 8.5, which is the same as 8 1/2 hours.

514.25/8.5 = 60.5

So the answer is 60.5 mph

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3 years ago

Evaluate log1∕6 36. please help

Sholpan [36]

$\log_{\tfrac 16} 36 \\\\\\=\dfrac{\ln 36}{ \ln \left(\tfrac 16\right)}\\\\\\=\dfrac{ \ln 6^2}{\ln (6^{-1})}\\\\\\=\dfrac{2 \ln 6}{-1 \ln 6}\\\\= - 2$

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3 years ago

What is the smallest number that rounds to 40

Liula [17]

Answer: 2

Step-by-step explanation:

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3 years ago

Read 2 more answers

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.