g The average salary in this city is $45,600. Is the average different for single people? 53 randomly selected single people who were surveyed had an average salary of $46,356 and a standard deviation of $15,930. What can be concluded at the α α = 0.05 level of significance?

Step-by-step explanation:

The average salary in this city is $45,600.

Using the formula

z score = x - u /(sd/√n)

Where x is 46,356, u is 45,600 sd is 15,930 and n is 53.

z = 46,356 - 45600 / (15930/√53)

z = 756/(15930/7.2801)

z = 756/(2188.1568)

z = 0.3455

To draw a conclusion, we have to determine the p value, at 0.05 level of significance for a two tailed test, the p value is 0.7297. The p value is higher than the significance level, thus we will fail to reject the null and can conclude that there is not enough statistical evidence to prove that the average is any different for single people.

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)

your answer is  

d.

196.5 m2

step-by-step explanation: