Five percent of workers in a city use public transportation to get to work. a transit authority offers discount rates to companes that have at least 30 employees who use public transportation to get to work. you randomly select 3 companies and ask their employees if they use public transportation below. complete parts (a) through (c).

(a) company a has 291 employees. what is the probability that company a will get the discount.

(b) company b has 540 employees. what is the probability that company b will get the discount.

(c) company c has 1013 employees. what is the probability that company c will get the discount.

a) the probability that Company A will get the discount is 0

b) the probability that Company B will get the discount is 0.2753

c) the probability that Company C will get the discount is 0.9986

Step-by-step explanation:

a) Let p(m) be the minimum proportion of workers needed at Company A who use public transportation to get to work in order the company can get discount offer by the transit authority.

Then ≈ 0.1031

The probability that Company A will get the discount can be stated as

P(z>z*) (p-value of z*) where z* is the z-score of 0.103 in the distribution of proportion of workers in the city who use public transportation to get to work.

z* can be calculated using the equation

where

Then z*= ≈ 4.15

P(z>4.15)=1-P(z<4.15)=1-1=0

b) for Company B:

≈ 0.0556

z*= ≈ 0.597

P(z>0.597)=1-P(z<0.597)= 1-0.7247=0.2753

c) for Company C:

≈ 0.0296

z*= ≈ -2.979

P(z>-2.979)=1-P(z<-2.979)=1-0.0014=0.9986

answer: '

the distance between points l and m

step-by-step explanation: