Answer:
0.1225
Step-by-step explanation:
Given
Number of Machines = 20
Defective Machines = 7
Required
Probability that two selected (with replacement) are defective.
The first step is to define an event that a machine will be defective.
Let M represent the selected machine sis defective.
P(M) = 7/20
Provided that the two selected machines are replaced;
The probability is calculated as thus
P(Both) = P(First Defect) * P(Second Defect)
From tge question, we understand that each selection is replaced before another selection is made.
This means that the probability of first selection and the probability of second selection are independent.
And as such;
P(First Defect) = P (Second Defect) = P(M) = 7/20
So;
P(Both) = P(First Defect) * P(Second Defect)
PBoth) = 7/20 * 7/20
P(Both) = 49/400
P(Both) = 0.1225
Hence, the probability that both choices will be defective machines is 0.1225
Answer:
A = 1500
3 years = $18,644.48
30 years = $132,032.78
Step-by-step explanation:
Step-by-step explanation:
I don't see options but here's the solution
2x²+9x-3=-x²+x
2x²+x²+9x-x-3=0
3x²+8x-3=0
(3x-1)(x+3)=0
Therefore x=⅓ or-3
AC
Same line segment just goes in the other direction.
<span>All the information we have are the probabilities, and what we need is the lowest number: so let's choose the smallest probability among the numbers: 0.0065%, B 0.0037%,C 0.0108%,D 0.0029%, E 0.0145%. The smallest of the numbers is 0.0029% -it starts with two 00s and the number that follows, 2, is smaller than all there others - so the smallest probability is in option D - and the model would be the corresponding model (but we're missing some information here) </span>
