Let X be the weekly incomes of a large group of executives. The weekly incomes of a large group of executives follows Normal distribution with mean $2000 and standard deviation $100.

μ =2000, σ =100

We have to find z score for income $2100 i.e x=2100

Z = $\frac{x-mean}{standard deviation}$

= $\frac{2100 - 2000}{100}$

Z = 100/100

Z = 1

The z score for income $2100 is 1

6 0

2 years ago

5u − –20u − u − 18u + –11u = 10<br> HURRRY PLEEEEESEEEE I ONLY HAVE 4 minutes left

drek231 [11]

5u - (-20u) - u - 18u + (-11u) = 10
5u + 20u - u - 18u - 11u = 10
-5u = 10
-5(-2) = 10

So, in conclusion, X, is equal to -2.

8 0

2 years ago

Qual o valor de 2 3 -1 3 ?

ra1l [238]

El valor de 23-13 es 10

7 0

2 years ago