Answer:
99.73% of bags contain between 62 and 86 chips .
Step-by-step explanation:
We are given that the number of chips in a bag is normally distributed with a mean of 74 and a standard deviation of 4.
Let X = percent of bags containing chips
So, X ~ N(
)
The standard normal z score distribution is given by;
Z =
~ N(0,1)
So, percent of bags contain between 62 and 86 chips is given by;
P(62 < X < 86) = P(X < 86) - P(X <= 62)
P(X < 86) = P(
<
) = P(Z < 3) = 0.99865 {using z table}
P(X <= 62) = P(
<=
) = P(Z <= -3) = 1 - P(Z < 3)= 1 - 0.99865 = 0.00135
So, P(62 < X < 86) = 0.99865 - 0.00135 = 0.9973 or 99.73%
Therefore, 99.73% of bags contain between 62 and 86 chips .
200 ÷ 5 = 40
4 • 40 = 160
James used 160 ft. of thread.
A. T(15) is the temperature of water after 15 seconds
C. T(60) is the temperature of the water after 1 minute
The average rate of change is
( f(6) - f(0) ) / (6-0) = (11/23 + 1/5) / 6 = 13/115
Sorry don’t now the answer to this problem