First we multiply 50 and 13 and 3/4
13 and 3/4 could also = 13.75
sooo lets multiply
50 x 13.75 = 687.50
so $687.50 was spent on the stock
now lets fund out how much they sold them for
we multiply 12 and 50
50 x 12 = 600
now we need to find out the loss by subtracting
$687.50 - $600 =$ 87.50
so that means this company lost $87.50 :(
hope this helps :)
and don't forget 2
MARK ME BRAINLIEST! :D
Answer:
1627190
Step-by-step explanation:
(see attached for reference)
Given the number 1627187, we can see that the number in the tens place is the number 8.
How we round this depends on the number immediately to the right of this number. (i.e the digit in the ones place)
Case 1: If the digit in the ones place is less less than 5, then the number in the tens place remains the same and replace all the digits to its right with zeros
Case 2: If the digit in the ones places is 5 or greater, then we increase the digit in the tens place and replace all the digits to its right with zeros.
In our case, the digit in the ones places is 7, this greater than 5, hence according to Case 2 above, we increase the digit in the tens place by one (from 8 to 9) and replace all the digits to its right by zeros giving us:
1627190
(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.
Answer:
Total amount of water = 5,200,000
Step-by-step explanation:
Given:
water produced = 50,000 quarts of water per week
Production drop = 5% = 0.05 per year
Number of week in year = 52 week
Find:
Total amount of water
Computation:
Sum = a / r
a = 50,000 x 52
a = 2,600,000
Sum = a / [1-r]
Sum = 2,600,000 / 5%
Sum = 2,600,000 / 0.05
Total amount of water = 5,200,000
