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Makovka662 [10]
3 years ago
5

When is it better to use a $20 off coupon? When is it better to use a 20% coupon? *

Mathematics
1 answer:
jasenka [17]3 years ago
6 0

It It better to use a $20 off coupon on small items, and a 20% off coupon on bigger items to get much more money off of your purchase

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The mean temperature for the first 7 days in January was 6 °C.
hodyreva [135]

Hope this help!!!

Have a nice day!!!

4 0
3 years ago
The volume of a cylinder with a height of 2 and radius of 3
prisoha [69]

Answer:

v = 18π

Step-by-step explanation:

v = πr²h

plug in the givens

v = π(3²)2

v = 18π   exact answer

V = 18 * 3.14 = 56.52    decimal approximation

6 0
3 years ago
In finance the notion of expected value is used to analyze investments for which the investor has an estimate of the chances ass
blagie [28]

Answer:

Step-by-step explanation:

The expected return is given as

Expected Return = SUM (Return i x Probability i). i=1,2,3.....

First investment

Probability of 0.7, it returns 60cents per dollars

Second investment

Probably of 0.3, it loses 20cents per dollar.

Expected return=(0.7×60)-(0.3×20)

Excepted return= 42-6

Excepted return=36cents

To dollars, 1cents is 0.01dollars

Then, 36cents = 0.36dollars

Expected return=$0.36

7 0
2 years ago
Find a particular solution to y" - y + y = 2 sin(3x)
leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

     =e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}

Multiplying both sides of equation (1) by integrating factor and integrating we get

\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I

I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx

8 0
3 years ago
Jami can mow acre 1/6 in 8 minutes If her rate is constant, can Jami mow 1 1/2 acres in one hour? Explain your reasoning
LenaWriter [7]
Time taken by Jami to mow 1/6 acres is 8 minutes=8/60=2/15 hours.
Rate at which Jami is mowing will be:
rate=(number of acres)/(time)=(1/6)/(2/15)
=1/6÷2/15
=1/6×15/2
=5/4 acres/hour
=1.25 acres/hour
From above calculations we conclude that Jami cannot mow 1.5 acres in 1 hour 

8 0
3 years ago
Read 2 more answers
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