A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 117.7-cm and a standard deviation of 2.2-cm. For shipment, 29 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 118.5-cm.

Required probability is 0.9748

Step-by-step explanation:

given data

mean = 117.7-cm

standard deviation = 2.2-cm

sample size n = 29

solution

we consider here random variable which represents here length of rod= x

so get here first z that is express as

put here value with x value 118.5-cm

Z = 1.9582

p value is 0.9748

so required probability is 0.9748

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3