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3 years ago

What's the cubic feet?

Mathematics

1 answer:

tino4ka555 [31]3 years ago

8 0

Answer:

126'

Step-by-step explanation:

V= l * w * h

V= 7 * 3 * 6

V= 126' which can be said to be 2064.77cm^3

You might be interested in

Which two lines are parallel?

ddd [48]

For lines to be parallel they must have the same slope. If:

y=mx+b

For any line to be parallel to the above it must also have the same value for m in that line.

4 0

3 years ago

Frank, who lives in Texas, and his sister Lilly, who lives in Japan, correspond regularly. From what he can tell from the postma

Hoochie [10]

Question:

The available options are:

(A) There is convincing evidence that there is no difference in the mean delivery times.

(B) There is convincing evidence that there is a difference in the mean delivery times.

(C) There is convincing evidence that the mean delivery time from Japan to Texas is greater than the mean delivery time from Texas to Japan.

(D) There is not convincing evidence that the mean delivery time from Japan to Texas is greater than the mean delivery time from Texas to Japan. (E) The t-test cannot be used for sample sizes that are this small.

Answer:

The correct option is;

(A) There is convincing evidence that there is no difference in the mean delivery times.

Step-by-step explanation:

Here we have the formula for the two-sample t-test given s;

$t =\frac{\overbar\overline{\rm x}_1 - \overbar\overline{\rm x}_2 }{\sqrt{\frac{\sigma ^2_1}{n_1} } +\frac{\sigma ^2_2}{n_2} }}$

To Texas N₁ = 12, $\overline{\rm x}_1$ =8.74 σ₁ = 2.92 SE Mean₁ = 0.84

To Japan N₂ = 9 $\overline{\rm x}_2$ = 6.75 σ₂ = 2.56 SE Mean₁ = 0.85

t = 1.667

Where the null hypothesis is

mu To Texas = mu To Japan and the alternative is

mu To Texas > mu To Japan

Here since the observed P value of P = 0.058 is greater than the significance level of 0.05 we fail to reject the null hypothesis and we will therefore, not accept the alternative hypothesis.

Hence, there is convincing evidence that there is no difference in the mean delivery times.

3 0

3 years ago

The length of pregnancy isn’t always the same. In pigs, the length of pregnancies varies according to a normal distribution with

Nesterboy [21]

a. percent of pig pregnancies that are longer than 106 days

Since we have a normal distribution here and the average number of days is 106, we can say that 50% of the pig pregnancies are longer than 106 days.

b. percent of pig pregnancies that are shorter than 111 days

To get the percentage, we will have to convert x = 111 days into a z-score first. The formula is:

$z=\frac{x-\mu}{\sigma}$

where x = raw data, μ = population mean, and σ = population SD.

Since these 3 pieces of information are already given in the question, let's plug them into the equation above.

$z=\frac{111-106}{5}=\frac{5}{5}=1\sigma$

Therefore, 111 days is located 1 standard deviation to the right of the mean.

To find the percentage of pig pregnancies shorter than 111 days, we need to find the area covered to the left of 1 SD.

To find the area covered to the left of 1SD, we need to use the standard normal distribution table.

Based on the table, the area covered to the left of 1SD is 0.8413. Multiplying the area by 100, we get 84.13. Therefore, 84.13% of the pig pregnancies are shorter than 111 days.