Answer:
(a)
(b)
Step-by-step explanation:
Alphaville's Budget Surplus Model is 
Betaville's Budget Surplus Model is 
We want to determine the expression that shows how much greater Alphaville’s annual budget surplus is than Betaville’s for a particular amount of tax revenue.
- To do this, we subtract Betaville's Model from Alphaville's model.
Opening the brackets
Collect like terms and simplify
The expression that shows how much greater Alphaville's Budget is:
(b) If the tax revenue that year in each town is $75,000
We want to evaluate the expression derived above when the tax revenue that year in each town is $75,000 i.e.at x=75000
Thtas what you must mean.
The distance beteen th epoints will be sqrt(4^2 + 6^2) = sqrt 52 = 2 sqrt13
Answer:
J Darren earns $17.35 per hour at his job. How many hours does he need to work in order to earn more than $624.60?
Step-by-step explanation:
You're looking for a situation in which the number 17.35 needs to be multiplied by an unknown quantity.
Scenarios F and G involve addition of 17.35 to something.
Scenario H would correspond to the inequality ...
17.65x < 624.60
The sign of the given inequality is >, so scenario H is not the one.
__
In scenario J, Darren's total earnings for x hours will be 17.35x, and we want to find x such that this total is greater than 624.60. This matches the given inequality:
17.35x > 624.60 . . . . . . answer choice J
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
