Help me plz on the question thx

HELP PLEASE WILL GIVE BRAINLIST TO BEST ANSWER!

Setler79 [48]

Answer:

12 cubic ft.

Step-by-step explanation:

A=bh=4(3)=12 ft^3

4 0

3 years ago

Read 2 more answers

Both before and after a recent earthquake, surveys were conducted asking voters which of the three candidates (Perez, Chung, Ste

Iteru [2.4K]

Step-by-step explanation:

We can notice that there is a change:

27⇒ 19
40⇒38
21⇒30

The votes for Perez and Chung decreased but the votes for Stevens increased

3 0

4 years ago

Marcy played the piano for 45 minutes. She stopped playing at 4:15 p.m. At what time did she start playing the piano ?

ololo11 [35]

Well, first off 1 hr =60 minutes. 60 min - 45 min = 15 minutes extra. what is one hour before 4:15? 3:15! Add the subtracted 15 minutes to get 3:30... Hope this helps

4 0

3 years ago

Young's modulus is a quantitative measure of stiffness of an elastic material. Suppose that for aluminum alloy sheets of a parti

neonofarm [45]

Answer:

a1) $\mu_{\bar{X}} = 70$

a2) $\sigma_{\bar{X}} = 0.4$

b1) $\mu_{\bar{X}} = 70$

b2) $\sigma_{\bar{X}} = 0.2$

c) X is more likely to be within 1 GPa of 70 GPa in the random sample of part b because of the largeness in sample size and less scattering of data

Step-by-step explanation:

Mean value, $\mu = 70$

Standard deviation, $\sigma = 1.6$

a1) sample size, n = 16

Mean of the sampling distribution of the sample mean = mean value, i.e.

$\mu_{\bar{X}} = \mu\\\mu_{\bar{X}} = 70$

a2) The standard deviation of the sampling distribution of the sample mean

$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n} } \\\sigma_{\bar{X}} = \frac{1.6}{\sqrt{16} }\\\sigma_{\bar{X}} = 0.4$

b1) For sample size, n = 64

Mean of the sampling distribution of the sample mean = mean value, i.e.

$\mu_{\bar{X}} = \mu\\\mu_{\bar{X}} = 70$

a2) The standard deviation of the sampling distribution of the sample mean

$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n} } \\\sigma_{\bar{X}} = \frac{1.6}{\sqrt{64} }\\\sigma_{\bar{X}} = 0.2$

c) X is more likely to be within 1 GPa of 70 GPa in the random sample of part b because it has a larger sample size, hence a decrease in the variability. This makes us easily determine the position of the sample around the population mean

5 0

4 years ago

The weights of bags filled by a machine are normally distributed with a standard deviation of 0.055 kilograms and a mean that ca

Sloan [31]

Answer:

Step-by-step explanation:

Given that the weights of bags filled by a machine are normally distributed with a standard deviation of 0.055 kilograms and a mean that can be set by the operator.

Let the mean be M.

Only 1% of the bags weigh less than 10.5 kilograms

i.e. P(X<10.5) = 0.01

corresponding Z value for P(Z<z) = 0.01 is -0.025

i.e. 10.5 = M-0.025(0.055)

Solve for M from the above equation

M = $10.5+0.025(0.055)\\=10.501375$

Rounding off we get

10.50 kgs

Mean weight should be fixed as 10.50 kg.

6 0

4 years ago