Find the missing side length

What’s the correct answer for this question?

Korvikt [17]

Answer:

1)

Volume of pyramid = 1/3(Ab)(h)

Where Ab is the area of base, h is height

Volume of cone = 1/3(Ab)(h)

a) Their formula for finding volume is same. Also, their painting heads are same.

b) Pyramids have a tetragonal base while cones have a polygonal base

2) Pyramids

Volume of cone = (1/3) πr²h

Since Area of a circle = πr²

So

Volume of pyramid = (1/3)(A)(h)

So we can use the formula of a circle in cone's formula

6 0

3 years ago

Read 2 more answers

Help me plz!!!!!!!!!!!!!!!!!!!!

inessss [21]

Answer:

14 2/3

Step-by-step explanation:

.....................

3 0

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:

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

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

$P(-1.2$

8 0

3 years ago

Multi step equation 7x+8=36

Vitek1552 [10]

Hey

Answer: 4

How do you find that out?

Divide the two numbers on the right, 36 and 8.

36/8=4

7 * 4 + 8 = 36

Hope this helped

4 0

3 years ago

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A software developer wants to know how many new computer games people buy each year. Assume a previous study found the standard

Marrrta [24]

Answer:

The minimum sample size required to ensure that the estimate has an error of at most 0.14 at the 95% level of confidence is n=567.

Step-by-step explanation:

We have to calculate the minimum sample size n needed to have a margin of error below 0.14.

The critical value of z for a 95% confidence interval is z=1.96.

To do that, we use the margin of error formula in function of n:

$MOE=\dfrac{z\cdot \sigma}{\sqrt{n}}\\\\\\n=\left(\dfrac{z\cdot \sigma}{MOE}\right)^2=\left(\dfrac{1.96\cdot 1.7}{0.14}\right)^2=(23.8)^2=566.42\approx 567$

The minimum sample size to have this margin of error is n = 567.

3 0

3 years ago