What is .375 as a fraction
2 answers:
Answer:
Step-by-step explanation:
To get rid of the decimal point in the numerator, we count the numbers after the decimal in 3.375, and multiply the numerator and denominator by 10 if it is 1 number, 100 if it is 2 numbers, 1000 if it is 3 numbers, and so on.
<em><u>Therefore, in this case we multiply the numerator and denominator by 1000 to get the following fraction:</u></em>
Then, we need to divide the numerator and denominator by the greatest common divisor (GCD) to simplify the fraction.
<em><u>The GCD of 3375 and 1000 is 125. When we divide the numerator and denominator by 125, we get the following: </u></em>
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The range of y=e^x+1 is y>1.
Answer:
b) No, it's not independent.
c) 0.02
d) 0.59
e) 0.57
f) 0.5616
Step-by-step explanation:
To answer this problem, a Venn diagram should be useful. The diagram with the information of Event 1 and Event 2 is shown below (I already added the information for the intersection but we're going to see how to get that information in the b) part of the problem)
Let's call A the event that she passes the first course, then P(A)=.73
Let's call B the event that she passes the second course, then P(B)=.66
Then P(A∪B) is the probability that she passes the first or the second course (at least one of them) is the given probability. P(A∪B)=.98
b) Is the event she passes one course independent of the event that she passes the other course?
Two events are independent when P(A∩B) = P(A) * P(B)
So far, we don't know P(A∩B), but we do know that for all events, the next formula is true:
P(A∪B) = P(A) + P(B) - P(A∩B)
We are going to solve for P (A∩B)
.98 = .73 + .66 - P(A∩B)
P(A∩B) =.73 + .66 - .98
P(A∩B) = .41
Now we will see if the formula for independent events is true
P(A∩B) = P(A) x P(B)
.41 = .73 x .66
.41 ≠.4818
Therefore, these two events are not independent.
c) The probability she does not pass either course, is 1 - the probability that she passes either one of the courses (P(A∪B) = .98)
1 - P(A∪B) = 1 - .98 = .02
d) The probability she doesn't pass both courses is 1 - the probability that she passes both of the courses P(A∩B)
1 - P(A∩B) = 1 -.41 = .59
e) The probability she passes exactly one course would be the probability that she passes either course minus the probability that she passes both courses.
P(A∪B) - P(A∩B) = .98 - .41 = .57
f) Given that she passes the first course, the probability she passes the second would be a conditional probability P(B|A)
P(B|A) = P(A∩B) / P(A)
P(B|A) = .41 / .73 = .5616
