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:
81,82,83
Step-by-step explanation:
What three consecutive integers have a sum of 246? Which means that the first number is 81, the second number is 81 + 1 and the third number is 81 + 2. Therefore, three consecutive integers that add up to 246 are 81, 82, and 83.
Answer:
Let x = software program
Let y = video game
x < 200 ; y < 300
x + y < 425
50x ; 35y
x = 200 ; y = 225
50(200) + 35(225) = 10,000 + 7,875 = 17,875
x = 125 ; y = 300
50(125) + 35(300) = 6,250 + 10,500 = 16,750
x = 175 ; y = 250
50(175) + 35(250) = 8,750 + 8,750 = 17,500
It is more profitable to maximize production of software program when working within the limits provided.
Step-by-step explanation:
The midrange of the data is 81 is your answer
Considering this is a reasonably solvable math problem, I will subtract 0.0058 from 36 and an answer choice should match.
36.0000 - 0.0058 = 35.9942
Your answer is 35.9942.
I hope this helps!
