Answer:
D
Step-by-step explanation:
none of the others make sense
THE ONE THAT YOU HAVE DONE IS CORRECT.
<em>ALL</em><em> </em><em>DIAGONAL</em><em> </em><em>MATRICES</em><em> </em><em>ARE</em><em> </em><em>SQUARE</em><em> </em><em>MATRIX</em><em>.</em><em>.</em><em>.</em><em>.</em>
<em>HOPE</em><em> </em><em>TH</em><em>I</em><em>S</em><em> </em><em>HELP</em><em>S</em><em> </em><em>YOU</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:

Step-by-step explanation:
If you expand an equation you get its equivalent expression
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
Hell naw, nobody is about to do that for you