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3 years ago

Which of the following tables does not represent a linear relationship between x and y?

Mathematics

1 answer:

vovangra [49]3 years ago

6 0

Answer:

The first one is not liner

Step-by-step explanation:

This isn't linear because the y values increase inequally. It adds 2 and then 3 instead of keeping a constant rate

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How to solve 5/6y + 1 = -1/2y + 1/4

True [87]

Answer:

y= -1/4

y= -0.25

Step-by-step explanation:

5/6y+1= -1/2y+1/4

Subtract 1 from both sides

5/6y= -1/2y-3/4

Add 1/2y to both sides

1 1/3y= -3/4

Divide both sides by 1/3

y= -1/4

y= -0.25

6 0

3 years ago

(-5) ÷ (-7) = _____ - 35 -35

MrRissso [65]

Answer:

0.714

Step-by-step explanation:

Wgat more explanation could I give you. lol

5 0

3 years ago

Read 2 more answers

I need HELP ASAP!!! !!!!!!!!!

gtnhenbr [62]

Round to the nearest percent

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3 years ago

Find the area of this parallelogram. 13 cm 15cm h 20cm

Margaret [11]

Answer:

260 square centimeters

Step-by-step explanation:

The formula for the area of a parallelogram is:

b x h

Where the base (b) is multiplied by the height (h) perpendicular to it.

The base here is 20 cm

The height perpendicular to the base is 13 cm

20 x 13 = 260

Or you can decompose and rearrange it into a rectangle. You will get 20 cm as the length and 13 cm as the width. The answer will be just the same!

Hope this helps!

7 0

3 years ago

Read 2 more answers

Let X be a random variable with probability mass function P(X = 1) = 1 2 , P(X = 2) = 1 3 , P(X = 5) = 1 6 (a) Find a function g

Goryan [66]

The question is incomplete. The complete question is :

Let X be a random variable with probability mass function

P(X =1) =1/2, P(X=2)=1/3, P(X=5)=1/6

(a) Find a function g such that E[g(X)]=1/3 ln(2) + 1/6 ln(5). You answer should give at least the values g(k) for all possible values of k of X, but you can also specify g on a larger set if possible.

(b) Let t be some real number. Find a function g such that E[g(X)] =1/2 e^t + 2/3 e^(2t) + 5/6 e^(5t)

Solution :

Given :

$P(X=1)=\frac{1}{2}, P(X=2)=\frac{1}{3}, P(X=5)=\frac{1}{6}$

a). We know :

$E[g(x)] = \sum g(x)p(x)$

So, $g(1).P(X=1) + g(2).P(X=2)+g(5).P(X=5) = \frac{1}{3} \ln (2) + \frac{1}{6} \ln(5)$

$g(1).\frac{1}{2} + g(2).\frac{1}{3}+g(5).\frac{1}{6} = \frac{1}{3} \ln (2) + \frac{1}{6} \ln (5)$

Therefore comparing both the sides,

$g(2) = \ln (2), g(5) = \ln(5), g(1) = 0 = \ln(1)$

$g(X) = \ln(x)$

Also, $g(1) =\ln(1)=0, g(2)= \ln(2) = 0.6931, g(5) = \ln(5) = 1.6094$

b).

We known that $E[g(x)] = \sum g(x)p(x)$

∴ $g(1).P(X=1) +g(2).P(X=2)+g(5).P(X=5) = \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$

$g(1).\frac{1}{2} +g(2).\frac{1}{3}+g(5).\frac{1}{6 }= \frac{1}{2}e^t+ \frac{2}{3}e^{2t}+ \frac{5}{6}e^{5t}$$

Therefore on comparing, we get

$g(1)=e^t, g(2)=2e^{2t}, g(5)=5e^{5t}$

∴ $g(X) = xe^{tx}$

7 0

3 years ago