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
heather can do 2 problems in 10 minutes....60 minutes in an hr
2/10 = x / 60...2 problems to 10 min = x problems to 60 min
10x = 120
x = 120/10
x = 12....so she can do 12 problems in 1 hr
Joel can complete 3 problems in 15 minutes...
3/15 = x / 60
15x = 180
x = 180/15
x = 12....so he can do 12 problems in 1 hr
so 24 problems can be done in 1 hr
Heather started with 3 problems done and Joel started with 2 problems done....added = 5 problems already done
and it is asking between 30 and 50 hrs.....so there is no equal sign in ur problem
ur answer is : 30 < 24x + 5 < 50 <==
To find the amount of time between two events, we can use subtraction, just like we do with regular subtraction problems. The subtraction will look something like this:
10:33
-
7:18
Like in a regular subtraction problem, we can subtract the digits that line up, like 10 and 7. Subtracting all of the corresponding digits, we get that it took Tony 3 hours and 15 minutes to finish baking the cookies.
0.5b+2=3
0.5b=3-2
0.5b=1
b=2
so brad purchased 2 pounds of bananas