Answer:34p+16g=208
47p+16g=208
Step-by-step explanation:
Let p=the price of each pound of peaches.
Let g=the price of each pound of grapes.
Each pound of peaches sells for $p, so 34 pounds of peaches will bring in 34p dollars. Each pound of grapes will bring in 45g dollars. Therefore, the total amount 34p+45g=192.25
Likewise, if 47 pounds of peaches and 16 pounds of grapes brought in $208 then
47p+16g=208
34p+45g=192.25
47+16g=208
<span>If Dingane has $8.00, and thirty percent of that money is from five cent coins, then 8 x 0.3 = $2.40 of Dingane's money is made of five cent coins. In this case the number of five cent coins is the number of cents divided by five: 240/5 = 48. Therefore, Dingane has forty-eight five-cent coins.</span>
Answer:
Sebastian didn't interpret the numbers correctly (See explanation)
Step-by-step explanation:
If you go up an elevator, you are gaining floors, so the number is positive.
If you go down an elevator, you are losing floors, so the number is positive.
Looking at the statement, he went 7 floors up to his room, then 9 floors down to the parking garage this can be represented as 
, since the 7 is positive and the 9 is negative.
Sebastian described the movement as 
, which is wrong - 9 should be -9 and -7 should be 7.
Hope this helped!
(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.
