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Gemiola [76]
3 years ago
6

Please help me I'm not sure on this one.

Mathematics
1 answer:
FrozenT [24]3 years ago
3 0

Answer:

  (a)  The y-intercept of Function A is greater

Step-by-step explanation:

The grid is not large enough to let you plot the points of Function A, but you can estimate where they might go.

The first point, (-9, -13), is off the bottom at the left side of the graph. It will be 3 units below the bottom line (-10) at the vertical x=-9. It is approximately (exactly) coincident with the graphed red line.

The second point, (6, 12), is off the top of the graph at the vertical x=6. That point is clearly above the red line.

These two points of Function A are sufficient to give you the idea that Function A is above Function B in the area of this graph.

Function A will have a greater y-intercept.

__

If you want to compute the y-intercept for Function A, you can start by computing the slope of the line. From the slope formula, we have ...

  m = (y2 -y1)/(x2 -x1)

  m = (12 -(-13))/(6 -(-9)) = 25/15 = 5/3

Then the y-intercept can be found from ...

  b = y1 -m·x1

  b = (-13) -(5/3)(-9) = -13 +15 = 2

The y-intercept for Function A is (0, 2), which is <em>greater than</em> the y-intercept of (0, -1) that Function B has.

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Answer:

a. P(x= at least 2)= 1- P(no pass)= 1- 0.27= 0.73

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c. Expected Value = mean = 2.35 which is more than C grade that is grade B

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Step-by-step explanation:

                       X      P(X)          X .P(X)         X²     X².P(X)

                        0       0.1                 0            0           0

                         1        0.17             0.17         1           0.17

                        2       0.21             0.42         4         0.84

                        3       0.32            0.96          9         2.88

<u>                          4       0.2               0.8           16       3.2  </u>

<u>∑                     10          1                   2.35         30     7.09</u>

Let X represent an event that the student has passed with at least a 2.

The probability of not passing (or below 2) is 0.1+ 0.17= 0.27

Then using law of complementation

a.  P(x= at least 2)= 1- P(no pass)= 1- 0.27= 0.73

b. Let Y represent the event that  a student has an A (4) given that he has passed the class with at least a C (2)

P(x)= 0.73

P(A)= 0.4

P(Y)= P(A)/P(X)= 0.4/0.73=0.55

c. Expected value= mean = 2.35 which is greater than C , Hence grade B

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d. Variance =  ∑X².P(X)   -  (∑X.P(X)²= 7.09- (2.35)² = 7.09- 5.5225=1.5675

 Standard Deviation = square root of Variance = √1.5675= 1.252

e. P(all pass) = P(A) +P(B) +P(C)= 0.2+ 0.32+ 0.21= 0.73

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