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noname [10]
3 years ago
5

The scores on an exam are normally distributed with a mean of 74 and a standard deviation of 7. What percent of the scores are g

reater than 81?
A. 50%

B. 13.5%

C. 84%

D. 16%
Mathematics
1 answer:
muminat3 years ago
3 0
The difference between 81 and the mean is 
   81 - 74 = 7
This is exactly the value of the standard deviation. You know that the "empirical rule" tells you 68% of all scores lie within 1 standard deviation of the mean. That tells you 32% of all scores lie beyond 1 standard deviation from the mean.

The normal distribution is symmetrical, so half of those (16%) lie above 1 standard deviation above the mean; the other half (16%) lie below 1 standard deviation below the mean. We're only concerned with the first group—those scores above 1 standard deviation above the mean.

The appropriate choice is
   D. 16%
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Answer:

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Step-by-step explanation:

8+4f=3f

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False

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The open circle means you should not include the number it is on.

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Please help it will help a lot i really need this!!
Lorico [155]

Answer:

A' = (0,0)

B'=(8,0)

C'=(8,2)

D'=(0,2)

It is not a rigid motion.

Step-by-step explanation:

To use the mapping rule, substitute the original x and y values in it.

The coordinates of A are (0,0).

Using the mapping rule, the x coordinate of A' = 2x0 = 0

Using the mapping rule, the y coordinate of A' = (1/2)x0=0

So A' will not change locations. The image will be at (0,0).

The coordinates of B are (4,0)

Using the mapping rule, the x-coordinate of B' = 2x4 = 8

Using the mapping rule, the y-coordinate of B' = (1/2)x0=0

Therefore the image of B' will be located at coorindate (8,0)

The coorindates of C are (4,4).

Using the mapping rule, the x-coordinate of C' = 2x4=8

Using the mapping rule, the y-coordinate of C' = (1/2)x4=2

Therefore the image of C' will be located at coordinate (8,2)

The coordinates of D are (0,4)

Using the mapping rule, the x-coordinate of D' = 2x0 = 0

Using the mapping rule, the y-coordinate of D' = (1/2)x4=2

Therefore the image of D' will be located at coordinate (0,2)

<em>Is the transformation a rigid motion?</em>

NO, this transformation is not a rigid motion because the relative distance between the points does not stay the same after they have been transformed. The transformation is not a translation , rotation, reflection, nor glide reflection.

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