Cube Root Of 923421 And 56696

Find the leg of each isosceles right triangle when the hypotenuse is of the given measure. Given = 8 cm.

kogti [31]

An isosceles right triangle is composed of a right angle and two congruent acute angles. In this case, 90 + 2x = 180 where x is the acute angle. x is equal to 45 degrees. Using sine law, 8/sin 90 = y / sin 45 where y is the leg of the triangle. the leg's length is equal to 4 sqrt 2.

5 0

2 years ago

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A survey of 35 people was conducted to compare their self-reported height to their actual height. The difference between reporte

Over [174]

Answer:

The test statistic is t = 3.36.

Step-by-step explanation:

You're testing the claim that the mean difference is greater than 0.7.

At the null hypothesis, we test if the mean difference is of 0.7 or less, that is:

$H_0: \mu \leq 0.7$

At the alternate hypothesis, we test if the mean difference is greater than 0.7, that is:

$H_1: \mu > 0.7$

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.

0.7 is tested at the null hypothesis:

This means that $\mu = 0.7$

A survey of 35 people was conducted to compare their self-reported height to their actual height.

This means that $n = 35$

From the sample, the mean difference was 0.95, with a standard deviation of 0.44.

This means that $X = 0.95, s = 0.44$

Calculate the test statistic

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{0.95 - 0.7}{\frac{0.44}{\sqrt{35}}}$

$t = 3.36$

The test statistic is t = 3.36.

3 0

3 years ago

Pls help me 10 points

AlladinOne [14]

Domain: (-6, 0, 3) Range: (-9, -7, 3, 5)

5 0

2 years ago

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Find the exact value of tan(165°) using a difference of two angles

Katyanochek1 [597]

Answer: $-2+\sqrt{3}$

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Work Shown:

Apply the following trig identity

$\tan(A - B) = \frac{\tan(A)-\tan(B)}{1+\tan(A)*\tan(B)}\\\\\tan(225 - 60) = \frac{\tan(225)-\tan(60)}{1+\tan(225)*\tan(60)}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+1*\sqrt{3}}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\$

Now let's rationalize the denominator

$\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\\tan(165) = \frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}\\\\\tan(165) = \frac{(1-\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{(1)^2-2*1*\sqrt{3}+(\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{1-2\sqrt{3}+3}{1-3}\\\\\tan(165) = \frac{4-2\sqrt{3}}{-2}\\\\\tan(165) = -2+\sqrt{3}\\\\$

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As confirmation, you can use the idea that if x = y, then x-y = 0. We'll have x = tan(165) and y = -2+sqrt(3). When computing x-y, your calculator should get fairly close to 0, if not get 0 itself.

Or you can note how

$\tan(165) \approx -0.267949\\\\-2+\sqrt{3} \approx -0.267949$

which helps us see that they are the same thing.

Further confirmation comes from WolframAlpha (see attached image). They decided to write the answer as $\sqrt{3}-2$ but it's the same as above.

5 0

2 years ago

What are the answers to the following?

hram777 [196]

90%=0.9
30%=0.3
115.9%=1.159
9%=0.09
7%=0.07

0.452=45.2%
0.006=0.6%
0.002=0.2%
0.05=5%
4.78=478%

5 0

2 years ago