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Vsevolod [243]
3 years ago
12

Ms. Parker determined that the mean test score in her first-period class

Mathematics
1 answer:
cluponka [151]3 years ago
6 0

Answer:

B. Period 1 had scores that were less spread out around the mean than

Period 2

Step-by-step explanation:

Because the standard deviation of Period 2's scores is greater than that of Period 1's, the test scores for Period 2 are going to be more spread out than Period 1's test scores. B is the most logical choice in this case.

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A'(- 4, 3 )

Under a counterclockwise rotation about the origin of 90°

a point (x, y ) → (- y, x )

A(3, 4 ) → A'(- 4, 3 )


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3 years ago
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Answer:

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2 years ago
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Ira Lisetskai [31]

Answer: Choice D) y = \pm \sqrt{x+5}

The steps to finding the inverse will have us swap x and y. Afterward, we solve for y

y = x^2 - 5 \\\\x = y^2 - 5 \\\\x+5 = y^2 \\\\y^2 = x+5 \\\\y = \pm \sqrt{x+5} \\\\

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

Extra info:

The inverse relation is not a function because of the plus minus. For instance, plugging x = 4 into y = \pm \sqrt{x+5} leads to y = -3 and y = 3 simultaneously. You would have to apply a domain restriction on y = x^2-5 to make it a one-to-one function, to make the inverse a function. One possible domain restriction is x > 0 which would lead to the inverse function y = \sqrt{x+5}

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