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monitta
3 years ago
15

Suppose that you plan on investing into a account paying simple interest. The formula for simple interest is I = Prt, where I is

the interest earned on a investment of P dollars, at the given rate r, over t years. If the banker tells you that the time for your investment is determine by the following t = (I)/r, would they be correct?
Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
3 0

Answer:

No

Step-by-step explanation:

No.

i = prt is correct; its result is the simple interest earned.  

If you want to solve for time, t, divide both sides by pr:

i/(pr) = t

You might be interested in
The 1984 movie footloose made $8,556,965 in its opening weekend. The 2011 remake of footloose made $15,556,113 in its opening we
mezya [45]

Answer:

  • absolute change: $6,999,148
  • relative change: 81.79%

Step-by-step explanation:

The absolute change is the difference between the earnings numbers:

  $15,556,113 -8,556,965 = $6,999,148

__

The relative change is the ratio of this difference to the original amount:

  $6,999,148/$8,556,965 × 100% ≈ 81.794748% ≈ 81.79%

6 0
3 years ago
There were 450 people at a play. The admission price was $2 for adults and $1 for children. The admission receipts were $680. Ho
vampirchik [111]

Answer:

Adults= 110

Children= 340

Step-by-step explanation:

Given data

let adults be x

and children be y

x+y= 450----------1

2x+y=680---------2

from 1

x= 450-y

put this in 2

2(450-y)=680

900-2y=680

900-680=2y

220=2y

y= 220/2

y= 110

x+110= 450

x= 450-110

x=340

4 0
3 years ago
One car costs $18,942.
oksian1 [2.3K]

Answer:

28,130

Step-by-step explanation:

18,942+9188=28,130

7 0
2 years ago
Read 2 more answers
Please help me :3 please explain
Ivahew [28]

Using a proportional relationship, the amounts are given as follows:

  • Tennis 3, Soccer 7.
  • Tennis 36, Soccer 84.

<h3>What is a proportional relationship?</h3>

A proportional relationship is a function in which the output variable is given by the input variable multiplied by a constant of proportionality, that is:

y = kx

In which k is the constant of proportionality.

For this problem, we have that:

  • The input variable is the number of tennis players.
  • The output variable is the number of soccer players.

From the first row of the table, the constant is given as follows:

k = 35/15 = 7/3.

Hence the relationship is:

y = 7/3x.

For the second row of the table, we have that x = 3, hence:

y = 7/3 x 3 = 7.

For the third row of the table, we have that y = 84, hence:

84 = 7/3x

x = 84 x 3/7

x = 36.

Then the amounts are given as follows:

  • Tennis 3, Soccer 7.
  • Tennis 36, Soccer 84.

More can be learned about proportional relationships at brainly.com/question/10424180

#SPJ1

7 0
1 year ago
"find the reduction formula for the integral" sin^n(18x)
dexar [7]
Let

I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx
For demonstration on how to tackle this sort of problem, I'll only work through the case where n is odd. We can write

\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx
\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx

For the remaining integral, we can integrate by parts, taking

u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax

\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx

For this next integral, we rewrite the integrand

\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax
\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)

So putting everything together, we found

I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx
I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)
I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax
\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax

\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax
7 0
3 years ago
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