Aheadline in a certain newspaper states that "most stay at first job less than 2 years." that headline is based on an online poll of 400 college graduates. among those polled, 78 % stayed at their first full-time job less than 2 years. complete parts (a) through (d). a. assuming that 50% is the true percentage of graduates who stay at their first job less than two years, find the mean and standard deviation of the numbers of such graduates in randomly selected groups of 400 graduates.

Mean, μ = 200

Standard deviation, σ = 10

Step-by-step explanation:

Data provided:

a) true percentage of graduates who stay at their first job less than two years

p = 50% = 0.5

q = 1 - p = 1 - 0.5 = 0.5

Number of graduates selected in a group, n = 400

Now,

The mean is calculated as:

Mean, μ = p × n

or

μ = 0.5 × 400 = 200

and,

standard deviation, σ =

or

σ =

or

σ = √100

or

σ = 10

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

~

a

step-by-step explanation:

u take $30 minus $24. the answer/original($30)