Canine Crunchies Inc. (CCI) sells bags of dog food to warehouse clubs. CCI uses an automatic filling process to fill the bags. Weights of the filled bags are approximately normally distributed with a mean of 45 kilograms and a standard deviation of 1.2 kilograms. Requried:
a. What is the probability that a filled bag will weigh less than 49.5 kilograms?
b. What is the probability that a randomly sampled filled bag will weight between 48.5 and 51 kilograms?
c. What is the minimum weight a bag of dog food could be and remain in the top 15% of all bags filled?
d. CCI is unable to adjust the mean of the filling process. However, it is able to adjust the standard deviation of the filling process. What would the standard deviation need to be so that no more than 2% of all filled bags weigh more than 52 kilograms?

your answer is 23/2 - 13.5

step-by-step explanation:

7/2 evaluates to

multiply x and 4.5

multiply x and 1

the x just gets copied along.

the answer is x

x

4.5*x evaluates to 4.5x

multiply 3 and 4.5x

multiply 1 and x

the x just gets copied along.

x

4.5*x*3 evaluates to 13.5x

7/2-4.5*x*3 evaluates to

  + 8 =

the answer is

7/2-4.5*x*3+8 evaluates to

answer: slight chance that it will happen

step-by-step explanation:

2

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)