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OLEGan [10]
3 years ago
11

Suppose the scores on an exam are normally distributed with a mean μ = 75 points, and standard deviation σ = 8 points. The instr

uctor wanted to "pass" anyone who scored above 69. What proportion of exams will have passing scores?
(a) .25 (b) .75 (c) .2266 (d) .7734 (e) -.75
Mathematics
1 answer:
galben [10]3 years ago
4 0

Answer:

(d):  0.7734

Step-by-step explanation:

Using the normcdf( statistical function built into my old TI 83-Plus calculator, I find that the area under the standard normal curve from the extreme left to the value 69 is 0.2266.  Thus, about 0.2266 of students would fail and

1.000-0.2266, or 0.7734 would pass.  This corresponds to Answer (d).

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A bag of gumballs contains 2 pink, 3 orange, 4 blue, 2 red, and 1 yellow
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2 years ago
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HELP ME PLEASE!!!
ser-zykov [4K]

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so we're really looking for a line whose slope is -2/3 and runs through 9,4.


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8 0
3 years ago
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Can i pls get help asap i will give brainly if the answer is right
Zanzabum
<h3>Answer:   0.5</h3>

This is equivalent to the fraction 1/2

==============================================================

Explanation:

The distance from A to B is 3 units. We can count out the spaces, or subtract the x coordinates of the two points and apply absolute value.

|A-B| = |-5-(-8)| = |-5+8| = |3| = 3

or

|B-A| = |-8-(-5)| = |-8+5| = |-3| = 3

We can say that segment AB is 3 units long.

--------------------------

The distance from A' to B' is 1.5 units because...

|A'-B'| = |-2.5-(-4)| = |-2.5+4| = |1.5| = 1.5

or

|B'-A'| = |-4-(-2.5)| = |-4+2.5| = |-1.5| = 1.5

The absolute values ensure the distance is never negative.

We can say A'B' = 1.5

---------------------------

Now divide the lengths of A'B' over AB to get the scale factor k

k = (A'B')/(AB)

k = (1.5)/(3)

k = 0.5

0.5 converts to the fraction 1/2.

The smaller rectangle A'B'C'D' has side lengths that are exactly 1/2 as long compared to the side lengths of ABCD.

3 0
3 years ago
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