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3 years ago

11

Using the equation ŷ = 2.6 − 3.1x, what is the predicted value if x is 4.3?

1 answer:

Salsk061 [2.6K]3 years ago

5 0

<h3>Answer: Choice A) -10.73</h3>

Work Shown:

Plug x = 4.3 into the equation given

ŷ = 2.6 − 3.1x

ŷ = 2.6 − 3.1(4.3)

ŷ = 2.6 − 13.33

ŷ = -10.73

Note: The symbol ŷ is read as "y hat". It's used to estimate the y value when it comes to regression equations.

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taurus [48]

Answer:

$f(x) = \frac{6}{x^2} -8$ or $f(x) = -\frac{6}{x^2} + 4$

Step-by-step explanation:

Given

$(x,y) = (1,-2)$ --- Point

$\int\limits^6_2 {(1 + 144x^{-6})} \, dx$

The arc length of a function on interval [a,b]: $\int\limits^b_a {(1 + f'(x^2))} \, dx$

By comparison:

$f'(x)^2 = 144x^{-6}$

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Take square root of both sides

$f'(x) =\± \sqrt{\frac{144}{x^6}}$

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Split:

$f'(x) = \frac{12}{x^3}$ or $f'(x) = -\frac{12}{x^3}$

To solve fo f(x), we make use of:

$f(x) = \int {f'(x) } \, dx$

For: $f'(x) = \frac{12}{x^3}$

$f(x) = \int {\frac{12}{x^3} } \, dx$

Integrate:

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We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = \frac{6}{1^2} + c$

$-2 = \frac{6}{1} + c$

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Make c the subject

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$f(x) = \frac{6}{x^2} + c$ becomes

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For: $f'(x) = -\frac{12}{x^3}$

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Integrate:

$f(x) = -\frac{12}{2x^2} + c$

$f(x) = -\frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = -\frac{6}{1^2} + c$

$-2 = -\frac{6}{1} + c$

$-2 = -6 + c$

Make c the subject

$c = -2+6$

$c = 4$

$f(x) = -\frac{6}{x^2} + c$ becomes

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Read 2 more answers