Given:
6 out of 2000 students quit community college due to serious health issues.
$10,000 insurance company offer
$60 per year
Based on the data, there are 6 students who will be given the $10,000 insurance per year. So they need
6 * 10,000 = $60,000 to make a break-even on their insurance offer
If a student pays $60 per year, they should have
$60,000 / $60 = 1,000 students who will pay per year to reach the break-even mark of their investment. If the number of students will exceed 1,000, the company will begin to earn.
In this question there are several important information's worth noting.Based on these information's the answer can be easily found. It is given that the total capacity of the tank is 24000 gallons. in 24 hours or in 1 day the amount of water used is 650 gallons. The question requires us to find the time it will take to use 2/3 rd of the water in the tank.
then
2/3 rd water in respect to total water in tank = (2/3) * 24000
= 16000
So 2/3 rd water of full tank is equivalent to 16000 gallons.
Now
650 gallons of water is used in = 24 hours
So
16000 gallons of water will be used in = (24/650) * 16000
= 590.77 hours
So it will take 590.77 hours to consume 2/3rd of the total water present in the tank.
Answer:
x > 36 in
Step-by-step explanation:
Let x = the width of the picture frame.
Then x + 6 = the length of the frame.
The formula for the perimeter P of a rectangle is'
P = 2l + 2w.
So, the condition is
2l + 2w > 156
2(x + 6) + 2x > 156 Distribute the 2
2x + 12 + 2x > 156 Combine like terms
4x + 12 > 156 Subtract 12 from each side
4x > 144 Divide each side by 4
x > 36
The perimeter of the picture frame will be greater than 156 in if x > 36 in.
(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.
10 1/2-4 1/4= 6 1/4. 6 1/4-3 2/3= 2 7/12. They used 2 7/12 ft or 2ft 7in of wood to fix the stairs.
