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
 
        
             
        
        
        
C(h,k) radius(r)
C(2,-3) r=4
(x-h)^2 + (y-k)^2 = r^2
(x-2)^2 + (y--3)^2 = 4^2
(x-2)^2 + (y+3)^2 = 16
therefore, the answer is D
        
                    
             
        
        
        
Y=-12
explanation multiply x by -3 to get y
        
                    
             
        
        
        
Domain and range of what? im not seeing a set of numbers id be happy to help tho
        
             
        
        
        
Answer:
-48/11
Step-by-step explanation:
-2x+5-40-9x-13=0
-11x=40-5+13
-11x=48
x=-48/11