The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 36 liters, and standard deviation of 3.9 liters.

a) what is the probability that daily production is between 37.4 and 37.5 liters? do not round until you get your your final answer.

answer= (round your answer to 4 decimal places.)

warning: do not use the z normal may not be accurate enough since wamap may look for more accuracy than comes from the table.

b mark me branliest thx

step-by-step explanation:

but which class are u may know that first

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)