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3 years ago

Find the balance in the account. $2,500 principal earning 4%, compounded quarterly, after 4 years

1 answer:

3 0

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2500\\ r=rate\to 4\%\to \frac{4}{100}\dotfill &0.04\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years\dotfill &4 \end{cases} \\\\\\ A=2500\left(1+\frac{0.04}{4}\right)^{4\cdot 4}\implies A=2500(1.01)^{16}\implies A\approx 2931.4466123$

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The Quality Control Department employs five technicians during the day shift. Listed below is the number of times each technicia

sveta [45]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data :

Technician __Shutdown

Taylor, T___4

Rousche, R _ 3

Hurley, H__ 3

Huang, Hu___2

Gupta, ___ 5

The Numbe of samples of 2 possible from the 5 technicians :

We use combination :

nCr = n! ÷ (n-r)!r!

5C2 = 5!(3!)2!

5C2 = (5*4)/2 = 10

POSSIBLE COMBINATIONS :

TR, TH, THu, TG, RH, RHu, RG , HHu, HG, HuG

Sample means :

TR = (4+3)/2 = 3.5

TH = (4+3)/2 = 3.5

THu = (4+2) = 6/2 = 3

TG = (4 + 5) = 9/2 = 4.5

RH = (3+3) = 6/2 = 3

RHu = (3+2) /2 = 2.5

RG = (3 + 5) = 8/2 = 4

HHu = (3+2) = 2.5

HG = (3+5) = 8/2 = 4

HuG = (2+5) / 2 = 3.5

Mean of sample mean (3.5+3.5+3+4.5+3+2.5+4+2.5+4+3.5) / 10 = 3.4

Population mean :

(4 + 3 + 3 + 2 + 5) / 5 = 17 /5 = 3.4

Population Mean and mean of sample means are the same.

This distribution should be approximately normal.

7 0

3 years ago

2. 3 qt =?L<br>15lb=?kg<br>7L=?qt

Firdavs [7]

3qt=2.83906L and 15lb=6.80389kg and 7L=7.39682qt hope this help

5 0

3 years ago

Read 2 more answers

What is the square root of <br> X of the power of 2 -1 =80?

zepelin [54]

Answer:

x=6561

Step-by-step explanation:

We can solve this equation for x to get

$\sqrt{x} -1=80\\\\\sqrt{x} =81\\x=(81)^2\\\\x=6561$

5 0

3 years ago

The area of the parallelogram is 273 square units. What is the height of the parallelogram? ____units

Alex17521 [72]

Answer:

h = 13 units²

Step-by-step explanation:

The way to find the area of a parallelogram is to multiply the length of one side (21) and multiply it by the height:

A = l*h

We know the area and the length:

273 = 21*h

Now we solve for height.

273/21 = (21*h)/21

273/21 = h

273/21 = 13

h = 13 units²

6 0

3 years ago

How do you solve his with working

AlexFokin [52]

Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

$\bf \begin{array}{cllll}
\textit{circumference of a circle}\\\\ 
2\pi r
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{arc's length}\\\\
s=\cfrac{\theta r\pi }{180}
\end{array}\\\\
-------------------------------$

$\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+
\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}
\\\\\\
15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888$

b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.

$\bf \begin{array}{cllll}
\textit{area of a circle}\\\\ 
\pi r^2
\end{array}\qquad \qquad \qquad \qquad 
\begin{array}{cllll}
\textit{area of a sector of a circle}\\\\
s=\cfrac{\theta r^2\pi }{360}
\end{array}\\\\
-------------------------------$

$\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}
\\\\\\
90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884$

7 0

3 years ago