Answer:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Step-by-step explanation:
For this case first we need to create the sample of size 20 for the following distribution:

And we can use the following code: rnorm(20,50,6) and we got this output:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Slope = (-6 - 3)/(1+2) = -9/3=-3
y - 3 = -3(x +2)
y - 3 = -3x - 6
3x + y = -3
answer is 3x + y = -3 (third one)
2 and 5. 1 isn't apart of the list because the definition of a prime number is having 1 and itself as a factor. 1 only has the number 1.
Use photomath that should give u the right answer too
Answer:
half of 8 is 4
so 4 is your radius
4^4 = 16
16 x 3.14 = 50.24 in2 is your area.