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Alexus [3.1K]
3 years ago
11

Multiplying fractions. Eight ninths times four fifths

Mathematics
1 answer:
Leni [432]3 years ago
6 0
\frac{8}{9} *\frac{4}{5} = \frac{8*4}{9*5}= \\\frac{32}{45} \frac{32}{45} [/tex]


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The passing yards for the top 5 quarterbacks in the country are 3,832, 3,779, 3,655, 3,642, and 3,579. Find the variance and sta
Leviafan [203]

First get the average passing yards:

(3,832 + 3,779 + 3,655 + 3,642 + 3,579) / 5 = 3,697.4

Now get the squared residuals for each quarterback's passing yards. That is, compute the difference between each data and the average, then square the result. For example,

(3,832 - 3,697.4)^2 = 134.6^2 = 18,117.2

For the others, you should get squared residuals of

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14,018.6

Take the sum of the squared residuals, then - since this is a sample, and not a population of all quarterbacks - divide the sum by 5 - 1 = 4 to get the variance:

(18,117.2 + 6,658.56 + 1,797.76 + 3,069.16 + 14,018.6)/4 = 10,915.3

The standard deviation is just the square root of the variance:

√(10,915.3) ≈ 104.48

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6 edges

I hope it helps

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A college student is taking two courses. The probability she passes the first course is 0.73. The probability she passes the sec
zhenek [66]

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

4 0
3 years ago
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