HELP PLEASE. Which regression equation best fits these data?

Mathematics

2 answers:

AlladinOne [14]2 years ago

7 0

Answer:

B. y = -0.58x^2 -0.43x +15.75

Step-by-step explanation:

The data has a shape roughly that of a parabola opening downward. So, you'll be looking for a 2nd-degree equation with a negative coefficient of x^2. There is only one of those, and its y-intercept (15.75) is in about the right place.

The second choice is appropriate.

_____

The other choices are ...

A. a parabola opening upward

C. an exponential function decaying toward zero on the right and tending toward infinity on the left

D. a line with negative slope (This might be a good linear regression model, but the 2nd-degree model is a better fit.)

creativ13 [48]2 years ago

5 0

Answer: y = -0.58x^2 -0.43x +15.75

Step-by-step explanation:

A P E X

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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Arada [10]

If $y=3x^2+30x+71$ , then
$y=3(x^2+10x)+71=3(x^2+2\cdot5x+25-25)+71$
$y=3((x+5)^2-25)+71=3(x+5)^2-75+71=3(x+5)^2-4$ . So, in the first gap you can write 5 and in the second you can write -4.
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1 year ago

Read 2 more answers

In the trapezoid ABCD ( AB ∥ CD ) point M∈ AD , so that AM:MD=3:5. Line l ∥ AB and going trough point M intersects diagonal AC a

geniusboy [140]

Answer:

BC:BN=8:3

Step-by-step explanation:

ABCD is a trapezoid and there is a point m which belongs to AD such that AM:MD=3:5.Line "l" parallel to AB intersects the diagonal AC at p and BD at N.

Now, we know that the parallel lines divide the transversal into the segments with equal ratio, therefore, BN:NC=AM:MD

But, BC= BN+NC

Therefore, BC:BN=(BN+NC):BN

⇒BC:BN=(3+5):3

⇒BC:BN=8:3

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2 years ago

A statue is mounted on top of a 16 foot hill. From the base of the hill to where you are standing is 77 feet and the statue subt

DanielleElmas [232]

Consider right triangle with vertices B - base of the hill, S - top of the statue and Y - you. In this triangle angle B is right and angle Y is 13.2°. If h is a height of the statue, then the legs YB and BS have lengths 77 ft and 16+h ft.

You have lengths of two legs and measure of one acute angle, then you can use tangent to find h:

$\tan 13.2^{\circ}=\dfrac{\text{opposite leg}}{\text{adjacent leg}} =\dfrac{16+h}{77}, \\ \\ 0.2345=\dfrac{16+h}{77},\\ \\ 16+h=0.2345\cdot 77=18.0565,\\ h=18.0565-16=2.0565$ ft.