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Lilit [14]
3 years ago
5

A student’s score on an exam is an integer between 0 and 100. A score of 90 to 100 is an A, 80 to 89 is a B, 70 to 79 is a C, 60

to 69 is a D, and below 60 is a failing grade of F. It is known that scores between 51 and 100 are equally likely and scores less than 50 cannot occur. Find the following probabilities: (a) Find the probability the student gets an A? (b) Find the probability that student fails the exam. (c) Find the probability that the student who is known not to have a C grade, has a B grade. (d) Find the probability that a student who is known to have passed the course, has a B grade.
Mathematics
1 answer:
elixir [45]3 years ago
4 0

Answer:

A. 0.22

B. 0.18

C. 0.25

D. 0.244

Step-by-step explanation:

S = {51 to 100} = 50

The sample space S contains values from 51 to 100 which is a total of 50 different values.

A.

Probability of A (lies between the values of 90 to 100 = 11).

11/50 = 0.22

B.

For a student to fail the course, his course has to be less than 60 = from 51 to 59. A total of 9 values.

9/50 = 0.18

C.

For student to get c, (70 to 79) a total of 10 values: 10/50 = 0.20

P(student did not get C) = 1-0.20 = 0.80

To get B, ( 80 to 89)

10/50 = 0.20

Probability that a student who is known not to have a c grade has a b grade = 0.20/0.80 = 0.25

D.

Probability of passing lies between 60 to 100 = 41 scores

41/50 = 0.82

Probability of student who passed having a B = 0.20/0.82 = 0.244

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alexandr402 [8]

Answer:

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Step-by-step explanation:

Your goal here is to REDUCE the given expression to simplest terms.

One way in which to approach this problem would be to rewrite (2x^2y^10)^3 as:  (2x^2*y^8)*y^2*(2x^2*y^10)^2.

Dividing this rewritten expression by 2x^2*y^8 results in:

y^2(2x^2*y^10)^2.

We now need to raise (2x^2*y^10) to the power 2.  Doing this, we get:

4x^4*y^20.

Multiply this by y^2 (see above):

y^2*4x^4*y^20

The first factor is 4:  4y^2*x^4*y^20.  This is followed by the product of y^2 and y^20:                   4*y^22*x^4

Finally, this should be re-written as

                                    4*x^4*y^22

Another way of doing this problem would involve expanding the numerator fully and then cancelling out like factors:

8*x^6*y^30      4*x^4*y^22

----------------- = ------------------ = 4*x^4*y^22

  2x^2y^8                  1

7 0
3 years ago
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Answer:

C

Step-by-step explanation:

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3 years ago
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3 years ago
1. 4x + 5y
kvv77 [185]
<h3><u>Solution</u><u> </u><u>:</u><u>-</u><u> </u></h3>

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1) 4x + 5y

→ 4*6 + 5*1

→ 24 +5 = 29

2) 2x (20-y)

→ 2*6 (20-1)

→ 12 *19 = 228

3) x ÷ 3 + y

→ 6 ÷ 3 + 1

→ 2 +1 = 3

4) 4y ÷ 2 + (8x + 10)

→ 4*1 ÷ 2 +(8*6+10)

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5 ) 5. Patty made $34 baby sitting on each of 3 weekends. If she spent $50 on gifts for her family, how much money does she have left?

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6.Carlos solved 20-(2x6) + 8 divide 4 = 29. Is this the correct answer? ( In attachment it's written 29 and not 28 .)

Given equation is ,

→ 20 - (2 × 6) + 8 ÷ 4 = 29

Solve inside the brackets ,

→ 20 - 12 + 8 ÷ 4

Perform division ,

→ 20 -12 + 2

Perform addition ,

→ 22 - 12

Perform subtraction,

→ 10

<u>No , he is wrong the answer should have been 10 but he got 29 .</u>

<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>

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