A pizza chain monitors the total weight of pepperoni that goes on its deluxe pepperoni pizzas to make sure customers are satisfied and product isn't being wasted. Suppose that for pizzas in this population, these weights are strongly skewed to the left with a mean of 250\,\text{g}250g250, start text, g, end text and a standard deviation of 8\,\text{g}8g8, start text, g, end text. Management takes a random sample of 646464 of these pizzas and calculates the mean weight of the pepperoni on the pizzas. Assume that the pizzas in the sample are independent.

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step-by-step explanation:

step-by-step explanation:

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)