Use the cosine function.
cos x = 2/12
cos x = 0.167
use the inverse of cosine:
x = 80.405
The answer is C.
Hope this helps :)
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:
6π cm³
Step-by-step explanation:
V = Ah ( A is the base area and h the height ), then
3πh = 18π ( divide both sides by 3π )
h = = 6
Then the height of the smaller cylinder = 6 - 3 = 3 cm
V = 3π × 3 = 9π cm³ ← volume of smaller can
Step-by-step explanation:
I’m not sure if this is correct but here is a picture of what I think the answer might be. P.S my handwriting is not neat