Milk and cream are mixed together for a recipe. The total volume of the mixture is 1 cup. If the milk contains 2% fat, the cream contains 18% fat, and the mixture contains 6% fat, how much cream is in the mixture? StartFraction 1 Over 5 EndFraction cup. cup StartFraction 1 Over 4 EndFraction cup. cup StartFraction 3 Over 4 EndFraction cup. cup StartFraction 4 Over 5 EndFraction cup. cup

1/4

Step-by-step explanation:

Let the volume of mixture (6%fat) in cup is 100cc

x  cc be the volume of cream (18% fat) in the mixture.

∴

(100−x) cc be the volume of milk (2% fat) in the mixture.

x * 0.18+(100−x) * 0.02=100

⋅0.06 or 0.18x − 0.02x = 6−2 or 0.16x

= 4 or x=25 cc  = 25%

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).