Answer:
The answer is below
Step-by-step explanation:
Let S denote syntax errors and L denote logic errors.
Given that P(S) = 36% = 0.36, P(L) = 47% = 0.47, P(S ∪ L) = 56% = 0.56
a) The probability a program contains both error types = P(S ∩ L)
The probability that the programs contains only syntax error = P(S ∩ L') = P(S ∪ L) - P(L) = 56% - 47% = 9%
The probability that the programs contains only logic error = P(S' ∩ L) = P(S ∪ L) - P(S) = 56% - 36% = 20%
P(S ∩ L) = P(S ∪ L) - [P(S ∩ L') + P(S' ∩ L)] =56% - (9% + 20%) = 56% - 29% = 27%
b) Probability a program contains neither error type= P(S ∪ L)' = 1 - P(S ∪ L) = 1 - 0.56 = 0.44
c) The probability a program has logic errors, but not syntax errors = P(S' ∩ L) = P(S ∪ L) - P(S) = 56% - 36% = 20%
d) The probability a program either has no syntax errors or has no logic errors = P(S ∪ L)' = 1 - P(S ∪ L) = 1 - 0.56 = 0.44
Answer:
-2 I believe. because the 1st exponent would make it 9, and then you'd remove 2
Step-by-step explanation:
I hope this is correct and helpful, I tried my best! <3
0.98 as a unsimplified fraction would be 98/100. The simplified fraction would be 49/50 hope this helped
12h + 30w represents the amount paid to each employee.....where h is the number of hrs and w is the number of wagons sold.
working 6 hrs and selling 3 wagons.....so h = 6 and w = 3
12h + 30w = total pay
12(6) + 30(3) =
72 + 90 =
162 = total pay <===
At least about 20 points it can depend on the class but not really.
