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vlada-n [284]
2 years ago
8

I need help with #4 a-d

Mathematics
1 answer:
lesya [120]2 years ago
6 0

a repeating decimal is placed over 9's instead of 0's.   for example: 0.23 = \frac{23}{100}, but 0.232323... = \frac{23}{99}

a) x = 0.121212

b) 100(x) = 100(0.121212)

    100x  = 12.1212

c)  100x = 12.1212

-  <u>        x =   0.1212</u>

     99x  =  12

         x  = \frac{12}{99}

d) -2\frac{12}{99}  <em>= -2\frac{4}{33} when simplified</em>

The number is rational becase it has a repeating decimal

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Answer:

Step-by-step explanation:

-4 + 13 - 4 + 13 - 4 + 13

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20. Based on student records, 25% of the students at a large high school have a
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Answer:

(E) 0.71

Step-by-step explanation:

Let's call A the event that a student has GPA of 3.5 or better, A' the event that a student has GPA lower than 3.5, B the event that a student is enrolled in at least one AP class and B' the event that a student is not taking any AP class.

So, the  probability that the student has a GPA lower than 3.5 and is not taking any AP  classes is calculated as:

P(A'∩B') = 1 - P(A∪B)

it means that the students that have a GPA lower than 3.5 and are not taking any AP  classes are the complement of the students that have a GPA of 3.5 of better or are enrolled in at least one AP class.

Therefore, P(A∪B) is equal to:

P(A∪B) = P(A) + P(B) - P(A∩B)

Where the probability P(A) that a student has GPA of 3.5 or better is 0.25, the probability P(B) that a student is enrolled in at least one AP class is 0.16 and the probability P(A∩B) that a student has a GPA of 3.5 or better and is enrolled  in at least one AP class is 0.12

So, P(A∪B) is equal to:

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∪B) = 0.25 + 0.16 - 0.12

P(A∪B) = 0.29

Finally, P(A'∩B') is equal to:

P(A'∩B') = 1 - P(A∪B)

P(A'∩B') = 1 - 0.29

P(A'∩B') = 0.71

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3 years ago
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