The product of 379 and 8 is 3,032.
It's between (any number less than 3,032) and (any number greater than 3,032).
I guess if this is an <em>estimation </em>exercise, you could say that it's between
(8 x 300) and (8 x 400), or 2,400 and 3,200.
Answer:
Part 1) m∠EFG=94°
Part 2) m∠GFH=86°
Step-by-step explanation:
we know that
m∠EFG+m∠GFH=180° -----> by linear pair (given problem)
we have
m∠EFG=3n+22
m∠GFH=2n+38
substitute the values
(3n+22)°+(2n+38)°=180°
Solve for n
(5n+60)=180
5n=180-60
5n=120
n=24
<em>Find the measure of angle EFG</em>
m∠EFG=3n+22
substitute the value of n
m∠EFG=3(24)+22=94°
<em>Find the measure of angle GFH</em>
m∠GFH=2n+38
substitute the value of n
m∠GFH=2(24)+38=86°
Answer: x = 6
<u>Step-by-step explanation:</u>
2[3x - (4x - 6)] = 5(x - 6)
2[3x - 4x + 6] = 5x - 30
2[-x + 6] = 5x - 30
-2x + 12 = 5x - 30
<u>+2x </u> <u>+2x </u>
12 = 7x - 30
<u>+30 </u> <u> +30 </u>
42 = 7x

6 = x
The area of a trapezpid is 1/2h×(b1+b2)
Answer:

Step-by-step explanation:
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The t distribution or Student’s t-distribution is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".
The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.
Data given
Confidence =0.99 or 99%
represent the significance level
n =16 represent the sample size
We don't know the population deviation 
Solution for the problem
For this case since we don't know the population deviation and our sample size is <30 we can't use the normal distribution. We neeed to use on this case the t distribution, first we need to calculate the degrees of freedom given by:

We know that
so then
and we can find on the t distribution with 15 degrees of freedom a value that accumulates 0.005 of the area on the left tail. We can use the following excel code to find it:
"=T.INV(0.005;15)" and we got
on this case since the distribution is symmetric we know that the other critical value is 