A car can travel 28 miles per 1 gallon of gas. How far can the car travel on 14 gallons of gas?

Which angles are pairs of corresponding angles? Check all that apply

AleksAgata [21]

Answer:

B and D people. Your welcome.

Step-by-step explanation:

6 0

3 years ago

WILL MARK BRAINLIEST Which equation can be used to find the volume of this solid?

AURORKA [14]

V=(7*6)/2 *8 the third option

you multiply the height with the base of the triangle and you get the area of the base of the prism so you multiply it with the height of the prism to get the volume

5 0

3 years ago

Find the value of x to the nearest tenth

TEA [102]

Answer:

114

Step-by-step explanation:

You would add 46 degree and 20 degrees (66) and subtract (180 - 66 = ?) to get your answer.

4 0

3 years ago

Read 2 more answers

The following prices, in dollars, of 7.5-cubic-foot refrigerators were recorded from a random sample. 314 305 344 283 285 310 3

Mariana [72]

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

3 years ago

ASAP (only the answer pls)

atroni [7]

Answer:

B..........................................

3 0

2 years ago