2. Suppose that the mean salary in a particular profession is $45,000 with a standard deviation of $1,500. What percentage of people in that profession earn less than $48,000

93%

Step-by-step explanation:

mean=45,000 standard deviation=2000 value of concern=48,000

We can easily see that since the value of concern (48,000) is GREATER than the mean, we can rule out the last two choices.

There is no possible way a number can be greater than the mean, but less than the 50th percentile.

convert 48,000 into a z-score, which is given as:

(x-mean)/standard deviation

or in this case:

(48000-45000)/2000=1.5

using my z-score table or calculator, I can see that a z-score of 1.5 corresponds to about the 93th percentile

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.