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3 years ago

The equation of the line of best fit of a scatter plot is y = 3x − 8. What is the slope of the equation?

Mathematics

2 answers:

alexira [117]3 years ago

7 0

It is the number (usually written to the left of x).
So the answer is 3

Irina-Kira [14]3 years ago

5 0

Y = 3x − 8. What is the slope of the equation?

C. 3

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Identify a3 of a sequence: 0.25, 0.5, 0.75, 1, 1.25, 1.5,....<br> a3=

wariber [46]

Answer:

0.75

Step-by-step explanation:

The given sequence is an AP , as the difference of all the terms is same

$a_3 = 0.75$

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4 years ago

Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.

NISA [10]

Answer:

$\frac{\sqrt{\pi}}{4}$

Step-by-step explanation:

You are going to integrate the following function:

$g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}$ (1)

furthermore, you know that:

$\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}$

lets call to this integral, the integral Io.

for a general form of I you have In:

$I_n=\int_0^{\infty}x^ne^{-ax^2}dx$

furthermore you use the fact that:

$I_n=-\frac{\partial I_{n-2}}{\partial a}$

by using this last expression in an iterative way you obtain the following:

$\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}}$ (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

$\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}$

5 0

3 years ago

Read 2 more answers

A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is

koban [17]

The expected length of code for one encoded symbol is

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha$

where $p_\alpha$ is the probability of picking the letter $\alpha$ , and $\ell_\alpha$ is the length of code needed to encode $\alpha$ . $p_\alpha$ is given to us, and we have

$\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}$

so that we expect a contribution of

$\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375$

bits to the code per encoded letter. For a string of length $n$ , we would then expect $E[L]=1.375n$ .

By definition of variance, we have

$\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2$

For a string consisting of one letter, we have

$\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4$

so that the variance for the length such a string is

$\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859$

"squared" bits per encoded letter. For a string of length $n$ , we would get $\mathrm{Var}[L]=1.859n$ .

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3 years ago

Mrs. Greensmith works a PC Richards and earns $35 per week plus a 12% commission o sales. How much would she earn for the week i

n200080 [17]

Answer: $245

she makes $35 every week (her wage, i’m assuming) regardless of how many sales she makes.

in order to find out how much money she makes out of her sales, you have to multiply the percentage commission by how much her sales was: so .12x1750 = $210 from sales.

you then add up her wage and her sales money, so $35 from wage + $210 from sales = $245 total for the week

answer: $245

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3 years ago

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50 POINTS<br><br> What is the volume of this?

yarga [219]

1468 is the answer.
You have to find the volume of the large shape and subtract the volume of the small shape. Write the equation as such:
(10*10*15) -(2*4*4)
1500 -32
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3 years ago

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