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3 years ago

Select the correct answer.

Mathematics

1 answer:

andrew-mc [135]3 years ago

7 0

Answer:

$v=\sqrt{\frac{E}{m}}$

Step-by-step explanation:

The formula is given as:

$E=mv^2$

We need to solve this formula for v, that means that v to one side and let it be solved in terms of the other variables (E and m). First, we isolate v:

$E=mv^2\\\frac{E}{m}=\frac{mv^2}{m}\\\frac{E}{m}=v^2$

To isolate v, and eliminate the "square", we need to take square roots of both sides, that will give us v in terms of the other variables:

$\frac{E}{m}=v^2\\\sqrt{\frac{E}{m}}=\sqrt{v^2}\\\sqrt{\frac{E}{m}}=v$

Putting v to left side (convention), we finally have:

$v=\sqrt{\frac{E}{m}}$

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For your answer x=13

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Each leg of a 45°-45°-90° triangle measures 12 cm.

vitfil [10]

Answer:

12 sqrt(2)

Step-by-step explanation:

Givens

a = 12 cm

b = 12 cm

c = ??

The triangle is isosceles with 1 right angle and 2 45 degree angles.

Formula

c^2 = a^2 + b^2

c^2 = 12^2 + 12^2

c^2 = 144 + 144

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c^2 = 144 * 2

c = sqrt(144) * sqrt(2)

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The answer isn't there. 12*sqrt(2) is the correct answer, 12sqrt(3).

Was this just a typo of some kind.

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3 years ago

Suppose that insurance companies did a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle

nirvana33 [79]

Answer:

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

And replacing we got:

$ME= 1.96*\sqrt{\frac{0.8 (1-0.8)}{400}} =0.0392$

ii) $Lower = 0.8-0.0392= 0.7608$

$Upper = 0.8 +0.0392= 0.8392$

Step-by-step explanation:

Notation and definitions

$X=320$ number of people that claimed always buckle up.

$n=400$ random sample taken

$\hat p=\frac{320}{400}=0.8$ estimated proportion of people that claimed always buckle up

$p$ true population proportion of people that claimed always buckle up

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

Part i

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: