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Katena32 [7]
2 years ago
6

A pyramid is placed inside a prism as shown. The pyramid has the same base area, B, as the prism but half the height, h, of the

prism. Which expression gives the volume of the pyramid?
Its not v=1/3*b*h

a.v=2/3bh
b. v=1/4bh
c.v=2bh
d.v=1/6bh
Mathematics
1 answer:
gayaneshka [121]2 years ago
7 0

Answer: D) V= 1/6 bh

Step-by-step explanation:

Since according to the given question,  A pyramid is placed inside a prism.

And, The pyramid has the same base area, b, as the prism but half the height, h, of the prism.

Therefore, base area of the pyramid = b

Height of the pyramid = h/2

Since, The volume of the pyramid,

V = 1/3 × base ares × height

V= 1/3 × b × h/2

V= 1/6 bh

Thus the required volume = 1/6 bh cube unit.

Therefore Option D is correct.



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Xander has 10 pieces of twine that he is using for a project. If each piece of twine is 1/3 yard long, what is the total length
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The following prices, in dollars, of 7.5-cubic-foot refrigerators were recorded from a random sample. 314 305 344 283 285 310​ 3
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Answer:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

Step-by-step explanation:

Information given

314 305 344 283 285 310​ 383​ 285​ 300​ 300

We can calculate the sample mean and deviation with the following formula:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

s =\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}

\bar X=310.9 represent the sample mean      

s=31.09 represent the standard deviation for the sample      

n=10 sample size      

\mu_o =300 represent the value to test

\alpha=0.05 represent the significance level

t would represent the statistic

p_v represent the p value

Hypothesis to test

We want to verify if the true mean is greater than 300, the system of hypothesis would be:      

Null hypothesis:\mu \leq 300      

Alternative hypothesis:\mu > 300      

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

Replacing the info given we got:

t=\frac{310.9-300}{\frac{31.09}{\sqrt{10}}}=1.109      

The degrees of freedom are given by:

df=n-1=10-1=9  

And the p value would be given by:

p_v =P(t_{9}>1.109)=0.148  

Since the p value is higher than the significance level of 0.05 we don't have enough evidence to conclude that the true mean is significantly higher than $300 because we FAIL to reject the null hypothesis.

6 0
2 years ago
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