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olchik [2.2K]
3 years ago
6

Finance. Suppose that $ 3 comma 400 is invested at 4.4 % annual interest​ rate, compounded monthly. How much money will be in th

e account in​ (A) 4 ​months? (B) 7 ​years?]
Mathematics
1 answer:
Yuki888 [10]3 years ago
3 0

Answer:

There would be $3,450.14 by the end of 4 months

There would be $4,623.78 by the end of 7 years.

Step-by-step explanation:

We are given the following in the question:

P = 3,400$

r = 4.4% = 0.044

Compounded monthly

Formula:

The compound interest is given by:

A = p\bigg(1+\dfrac{r}{n}\bigg)^{nt}

where A is the amount, p is the principal, r is the interest rate, t is the time in years and n is the nature of compound interest.

a) 4 months

A = 3400\bigg(1+\dfrac{0.044}{12}\bigg)^{4}\\\\A = \$3,450.14

There would be $3,450.14 by the end of 4 months.

b)  7 ​years

t = 7

A = 3400\bigg(1+\dfrac{0.044}{12}\bigg)^{84}\\\\A = \$4,623.78

There would be $4,623.78 by the end of 7 years.

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just add all the sides. since it’s a rectangle, you there is 2 sides that’s 3ft long and that are 6ft long. so you would do 3+3+6+6

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25 times a letter Y = a number 5
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The hover dam contains 3 1/2 millon cubic yards of concrete.The grand coulee dam contains 2 1/2 times as much. How much concrete
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grand coulee dam (g) contains 2.5 times as much concrete as the hover dam (h)

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Find the equation of a line that is perpendicular to y = 3x – 5 and passes through the point (1, -3).
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keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5

well then therefore

\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y+3=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{1}{3}-3\implies y=-\cfrac{1}{3}x-\cfrac{8}{3}

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