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svetlana [45]
1 year ago
5

When we use a least-squares line to predict y values for x values beyond the range of x values found in the data, are we extrapo

lating or interpolating? are there any concerns about such predictions?.
Mathematics
1 answer:
BabaBlast [244]1 year ago
6 0

Option A is correct that is we are extrapolating. Because the pattern of the data may vary outside the x range, extrapolation is risky.

Given that,

We have to find are we extrapolating or interpolating when we use a least-squares line to forecast y values for x values outside the range of x values seen in the data.

Option A) We are extrapolating. Because the pattern of the data may vary outside the x range, extrapolation is risky.

<h3>What is extrapolation?</h3>

Extrapolation is the process of estimating, beyond the original observation range, the value of a variable on the basis of its relationship with another variable.

Therefore, Option A is correct that is we are extrapolating. Because the pattern of the data may vary outside the x range, extrapolation is risky.

To learn more about extrapolating visit: brainly.com/question/14643498

#SPJ4

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What is (-4,5) and (8,-1)
Musya8 [376]

Answer:

the slope-intercept equation of a line that passes through the coordinate (-4,5) and (8,-1) is

y  =  - 0.5x + 3

Step-by-step explanation:

1) Apply the slope formula; y^2 - y^1 divided by x^2 - x^1

y^2 is -1

y^1 is 5

x^2 is 8

x^1 is -4

-1 - 5 = -6

8 - (-4) = 12

-6/12 is -1/2 or -0.5

2) To find the y-intercept, hoose either of the coordinates and replace y, m, and x in the y= mx+b formula.

5= -0.5(-4) + b

5= 2 + b

3= b

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3 years ago
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Step-by-step explanation:

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4 years ago
Find the domain of f(x) = x+5/ 2-x
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I don’t know I am stuck on thisss
8 0
2 years ago
A snack mix recipe calls for 2 1/5 cups of grapes and 1/4 cup of cheese. Skyler wants to make the same recipe using 1 cup of che
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4 0
3 years ago
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Solve for w, where w is a real number.
Ber [7]

Answer:

w= 9

Step-by-step explanation:

\sqrt{ - 4w + 61}  = w - 4

Square both sides:

-4w +61= (w -4)²

\boxed{(a - b)^{2}  = a^2 -2ab + b^2 }

Expand:

-4w +61= w² -2(w)(4) +4²

-4w +61= w² -8w +16

Simplify:

w² -8w +16 +4w -61= 0

w² -4w -45= 0

Factorize:

(w -9)(w +5)= 0

w -9= 0 or w +5= 0

w= 9 or w= -5 (reject)

Note:

-5 is rejected since we are only taking the positive value of the square root here. If the negative square root is taken into consideration, then w= -5 would give us -9 on both sides of the equation.

<u>Why </u><u>do </u><u>we </u><u>use </u><u>negative </u><u>square </u><u>root?</u>

When solving an equation such as x²= 4,

we have to consider than squaring any number removes the negative sign i.e., the result of a squared number is always positive.

In the case of x²= 4, x can be 2 or -2. Thus, whenever we introduce a square root, a '±' must be used.

However, back to our question, we did not introduce the square root so we should not consider the negative square root value.

6 0
2 years ago
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