My answer -
3 2/3
Wow you have allot of cats LOL!!!
P.S
Have an AWESOME LGBT DAY !!! :)
Not sure however I keep getting 1
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
Area of a circle = pi x r^2
200.96 = pi x r^2
200.96 = 3.14 x r^2
r^2 = 64
r = 8 inches
Hope this helps!! :)
Answer:
D) {-5,1}
R) {-4,7}
x (-31/11,0)
y (0,31/6)
<em><u>Hope this helped! Have a nice day! Plz mark as brainliest!!! :D</u></em>
<em><u>-XxDeathshotxX</u></em>