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3 years ago

Bradley does lawn work on the weekends. He charges $12 per lawn and $5 per hour

Mathematics

1 answer:

RoseWind [281]3 years ago

3 0

Answer:

The answer would be 'D'. 2 x 12 + 5h

Step-by-step explanation:

In this case, you know that Bradley mows two lawns and for every lawn, he charges $12. If you were not given the number of lawns mowed, you would've had something like 12x with 12 representing the charge for every lawn mowed and x representing the actual amount of lawns mowed. However, since you were given the amount of lawned mowed (2) you can put 12(2) or 12 x 2 instead of 12x.

With this, you can eliminate every answer that does not include 12 x2 or 2 x 12. (You can eliminate answers A and B)

Now for the number of hours for the bushes trimmed. The same rules apply to this.

The variable (h) would be included here since we were not given the information on how many hours it took. 5 represents the amount that he charges for every hour and h is the unknown amount. However, if we were also given the info for this like let's say 3 hours, it would be 5(3) instead of 5x. But that's a different story.

So with this let's combine the two terms. You have 2 x 12 and 5h so, in the end, you have: 2 × 12 + 5h

I hope my explanation wasn't too confusing. I tried my best to put it in simple terms

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Find the slope of the line passing through the points (-6,2), (0,-6).A 6B 2.C-4/3DOE-3/1

NNADVOKAT [17]

Given data:

The first point given iis (a, b)=(-6,2).

The second point given is (c,d)=(0, -6).

The expression for the slope is,

m=(d-b)/(c-a)

Substitute the given points in the above expression.

m=(-6-2)/(0-(-6))

=(-8)/(6)

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Thus, the slope of the line is -4/3, so (C) option is correct.

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1 year ago

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LenaWriter [7]

Answer:

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Step-by-step explanation:

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$\phi(t) = E[e^{tX}]$

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And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

$E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]$

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$\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}$

And then we have this:

$E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...$

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And we can find the variance like this :

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blondinia [14]

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prisoha [69]

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Step-by-step explanation:

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