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3 years ago

How do you figure out the height of the triangular pyramid??

Mathematics

1 answer:

miv72 [106K]3 years ago

8 0

We know that formular for volume of pyramid=area of base ×height
the area of the base is 1/2×base×height
=1/2×15×14
=105cm²
the given volume is 270cm³
∴height=volume÷base area
=270cm³÷105cm²
= $\frac{18}{7}$
= $2 \frac{4}{7}$

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How to use the prime factorization and the distributive property to express the sum of 2 whole numbers as a product

likoan [24]

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12= 2x2x3

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3 years ago

Molly is raising cows. She has a pasture for them that is 0.65 square miles in area. One dimension of the pasture is 0.4 miles l

Gala2k [10]

Answer:

1.625 miles.

Step-by-step explanation:

Molly is raising cows. She has a pasture for them that is 0.65 square miles in area. One dimension of the pasture is 0.4 miles long.

We have to find the length of the other side of the pasture.

Now, area if the pasture = product of it's two dimensions.

So, 0.65 = 0.4L

⇒ L = 1.625 miles. (Answer)

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3 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

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4 years ago

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strojnjashka [21]

Answer:

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Step-by-step explanation:

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2 years ago