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yaroslaw [1]
3 years ago
7

Is the volume of a triangular prism one-third the volume of a rectangular prism with the same base area and height?

Mathematics
1 answer:
Vlad [161]3 years ago
3 0

Nope

Step-by-step explanation:

Volume of rectangular prism :

= Base Area × h = l × w × h

Volume of triangular prism :

= Base Area × h

Cuz the base area of triangular prism, the volume is :

(½ × l × w) × h

= ½ lwh

So, the volume of a triangular prism is half of a rectangular prism

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B and C; as the y intercept increases/decreases, the graph of the line shifts up/down.

The y intercept is where the function crosses the y-axis. If the y intercept moves either up or down, the whole function will be translated up/down vertically along the y-axis.
5 0
2 years ago
Can you help me with this question?
Rufina [12.5K]

Answer:

  16.2

Step-by-step explanation:

The angle internal to the triangle at B is the supplement of the one shown, so is 65°. That is equal to the angle internal to the triangle at D. Since the vertical angles at C are congruent, the two triangles are similar by the AA theorem.

Corresponding sides of similar triangles are proportional, so we can write the proportion shown in the attachment:

  BC/FC = DC/AC

  BC = FC(DC/AC) = 21.6(7.2/9.6)

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3 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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Answer: y=1/2x-17

Step-by-step explanation:

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