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3 years ago

Which equation is represented by the table?

1 answer:

6 0

(0,-1)(1,4)
slope = (4 - (-1) / (1 - 0) = (4 + 1)/1 = 5

y = mx + b
slope(m) = 5
(1,4)...x = 1 and y = 4
now we sub
4 = 5(1) + b
4 = 5 + b
4 - 5 = b
-1 = b

equation is : y = 5x - 1...answer C

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F(x) = 3x + 1; g(x) = 5x - 1<br> Find f/g. (1 point)

zysi [14]

Answer:

$\frac{f}{g} =\frac{3x+1}{5x-1}$

$x\neq \frac{1}{5}$

Step-by-step explanation:

we are given

$f(x)=3x+1$

$g(x)=5x-1$

we have to find f/g

we can write as

$\frac{f}{g} =\frac{f(x)}{g(x)}$

now, we can plug values

and we get

$\frac{f(x)}{g(x)} =\frac{3x+1}{5x-1}$

so,

$\frac{f}{g} =\frac{3x+1}{5x-1}$

we know that denominator can not be zero

so,

$5x-1\neq 0$

now, we can solve for x

$5x-1+1\neq 0+1$

$5x\neq 1$

Divide both sides by 5

and we get

$x\neq \frac{1}{5}$

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3 years ago

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Lubov Fominskaja [6]

Answer:

4,400,073

Step-by-step explanation:

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2 years ago

5 feet 8 inches equals how many inches

sweet [91]

There are 12 inches in a foot. To find how many inches are in 5 feet, multiply by 12:

5*12 = 60 inches

Add the 8 additional inches:

60 + 8 = 68 inches

The answer is 68.

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3 years ago

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using the equation G = 10-1.25t. What is the x-intercept of the equation and what is its interpretation in the context of the pr

Masja [62]

Answer:

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3 years ago

A population has a standard deviation of 5.5. What is the standard error of the sampling distribution if the sample size is 81?

VladimirAG [237]

Answer:

$\sigma = 5.5$

And we have a sample size of n =81. We want to estimate the standard error of the sampling distribution $\bar X$ and for this case we know that the distribution is given by:

$\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$\sigma_{\bar x}= \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$\sigma_{\bar x}=\frac{5.5}{\sqrt{81}}= 0.611$

Step-by-step explanation:

For this case we know the population deviation given by:

$\sigma = 5.5$

And we have a sample size of n =81. We want to estimate the standard error of the sampling distribution $\bar X$ and for this case we know that the distribution is given by:

$\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$\sigma_{\bar x}= \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$\sigma_{\bar x}=\frac{5.5}{\sqrt{81}}= 0.611$

4 0

3 years ago