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
Answer:
Number of sand bags needed = 124
Step-by-step explanation:
Diameter of garden = 40 yd
Width of path = 6 yd
Diameter of garden with path = 40 + 2 x 6 = 52 yd
We need to find area of path.
Area of path = Area of garden with path - area of garden

Area covered by one sand bag = 7 yd²

Number of sand bags needed = 124
Answer:
Call the numbers a, b, c, d. What does it say if their mean is 15? That would be that added together then divided by 4 you get 15. So the number before you divided by 4, which is the sum, was 60. If you know some algebra, the equation (a+b+c+d)/4=15 gives the sum of 60 when you multiply both sides
B. 4(6x) and 24x.
Hope this helps!