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MArishka [77]
4 years ago
13

A triangular prism is shown below. the base of the prism is a right triangle.

Mathematics
1 answer:
Nimfa-mama [501]4 years ago
6 0
The product is 60.

We start out with the information we have.  This forms a right triangle, with one leg being 24, the other leg x, and the hypotenuse 30.  Using the Pythagorean theorem we have

x²+24²=30²
x²+576=900

Subtract 576 from both sides:
x²+576-576 = 900-576
x²=324

Take the square root of each side:
√x²=√324
x=18

We know that the formula for the volume of a triangular prism is V=(1/2bh)H, where b is the base of the triangle, h is the height of the triangle, and H is the height of the prism. Substituting our known information we have:

720=(1/2*18*24)y
720=216y

Divide both sides by 216:
720/216 = 216y/216
3 1/3 = y

This means the product of x and y would be:
18(3 1/3)
18(10/3)
180/3
60
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Answer:

hey hope this helps

<h3 /><h3>Comparing sides AB and DE </h3>

AB =

\sqrt{ {1}^{2} +  {1}^{2}  }

=  \sqrt{2}

DE

= \sqrt{ {(3 - 5)}^{2} +  {(1 + 1}^{2}  }  \\   = \sqrt{ {( - 2)}^{2}  +  {(2)}^{2} }  \\    = \sqrt{4 + 4}  \\  =  \sqrt{8}  \\   = 2 \sqrt{2}

So DE = 2 × AB

and since the new triangle formed is similar to the original one, their side ratio will be same for all sides.

<u>scale factor</u> = AB/DE

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It's been reflected across the Y-axis

<em>moved thru the translation of 3 units towards the right of positive x- axis </em>

for this let's compare the location of points B and D

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7 0
2 years ago
Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o
kirza4 [7]

Answer:

a) H0: \sigma = 1.34

H1: \sigma \neq 1.34

b) df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

c) t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

\sigma_o =1.34 the value that we want to test

p_v represent the p value for the test

t represent the statistic  (chi square test)

\alpha=0.01 significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: \sigma = 1.34

H1: \sigma \neq 1.34

The statistic is given by:

t=(n-1) [\frac{s}{\sigma_o}]^2

Part b

The degrees of freedom are given by:

df = n-1= 10-1=9

And the critical values with \alpha/2=0.005 on each tail are:

\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589

Part c

Replacing the info we got:

t=(10-1) [\frac{1.186}{1.34}]^2 =7.05

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

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3 years ago
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aleksandr82 [10.1K]

Answer: x = -1 and x = 1

Step-by-step explanation:

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Therefore : the solution to f(x) = g(x) is the point where x = -1 and x = 1

5 0
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