The workers' union at a particular university is quite strong. about 96% of all workers employed by the university belong to the workers' union. recently, the workers went on strike, and now a local tv station plans to interview 4 workers (chosen at random) at the university to get their opinions on the strike. what is the probability that exactly 3 of the workers interviewed are union members?

0.14155776

Step-by-step explanation:

Given : The proportion of workers employed by the university belong to the workers' union=0.96

Let x be a binomial variable that represents that worker belongs to the workers' union.

Sample size : n= 4

[Binomial probability formula : P(X=x)=^nC_x(p)^x(1-p)^{n-x}]

Hence, the probability that exactly 3 of the workers interviewed are union members

the inequality is:

250c+180g≥950

step-by-step explanation:

let c denotes the number of days in a week   miyoko worked as a cryptographer.

and g denotes the number of days in a week she works as a geologist.

now we are given that she earns $250 for every day she works as a cryptographer,and earns $180 for every day she works as a geologist.

now miyoko's goal is to earn more than $950 this week.

hence, the inequality could be represented as:

250c+180g≥950

1/3

step-by-step explanation:

to find this, you can divide 12 by 4 which is 3 or you can multiply 4 until you reach 12 (4x3=12). both should give you 3, meaning theres three times less bananas than there are oranges.