Hey! please help i’ll give brainliest! if it lets me:)

A small refrigerator is a cube with a side length of 12 inches use the formula S= 6s2 to find the surface area of the cube

Luda [366]

Answer:

Surface area of the refrigerator is 864 in².

Step-by-step explanation:

Formula to calculate the surface area of a cube is,

Surface area = 6s²

Here, s = length of a side of the cube

If side length 's' = 12 inches

By substituting this value in the formula,

Surface area = 6(12)²

= 6 × 144

= 864 inch²

Therefore, surface area of the refrigerator is 864 in².

6 0

3 years ago

What are 3 different quotients that equal 4 to the negative 5th power

Ilia_Sergeevich [38]

4^-5=1/1024 or 2/2048 or 3/3072

8 0

3 years ago

What goes up and down without moving?<br> Whoever answers first gets brainliest.

pav-90 [236]

Answer: temperature and weight

Step-by-step explanation: Temperature can increase or decrease/can go up and down without moving. The same goes with weight, it can go up or down without moving.

4 0

3 years ago

mackenzie made 33 out of 50 throws. what is the experimental probability that mackenzie will not make the next free throw she at

siniylev [52]

answer:

1.94117647058

Step-by-step explanation:

33/50 her free throw % is 66% at the moment based on how many she missed you'd do 33 / by 17 you get 1.94117647058

5 0

3 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

3 years ago