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gavmur [86]
3 years ago
15

If you open an account that compounds quarterly in 2020. In what year will it double the starting amount if it has an APR of 3.2

%
Mathematics
1 answer:
Lerok [7]3 years ago
8 0

Answer:

It will double in the year 2063

Step-by-step explanation:

Let the amount deposited be $x, when it doubles, the amount becomes $2x

we can use the compound interest formula to know when this will happen

The compound interest formula is as follows;

A = P(1+r/n)^nt

In this question,

A is the amount which is 2 times the principal and this is $2x

P is called the principal and it is the amount deposited which is $x

r is the interest rate which is 3.2% = 3.2/100 = 0.032

n is the number of times compounding takes place per year which is quarterly which equals to 4

t is the number of years which we want to calculate.

Substituting all these into the equation, we have;

2x = x(1+0.032/4)^4t

divide through by x

2 = (1+ 0.008)^4t

2 = (1.008)^4t

we use logarithm here

Take log of both sides

log 2 = log (1.008)^2t

log 2 = 2t log 1.008

2t = log 2/log 1.008

2t = 86.98

t = 86.98/2

t =43.49 which is 43 years approximately

Thus the year the money will double will be 2020 + 43 years = 2063

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There are 37 children in the classroom, each student will get 48 pencils.How many pencils will the teacher have to give out?
givi [52]

Answer:

1,776

Step-by-step explanation:

1. There are <u>37</u> children in the classroom

2. Each student will get <u>48</u> pencils

3. The teacher wants to give 48 pencils evenly to 37 children

4. 48 x 37 = <u>1,776</u>

A: The teacher will have to give out <u>1,776</u> pencils.

6 0
2 years ago
Kirk's average driving speed is 7 kilometers per hour faster than Thor's. In the same length of time it takes Kirk to drive 450
alexandr402 [8]

Answer:

x = 816 kilometers per hour.

Step-by-step explanation:

Given the following data;

Distance traveled by Kirk = 450km

Distance traveled by Thor = 408 km

Let the average driving speed of Thor be x.

Translating the word problem into an algebraic equation;

k = x + 6

We know that the time taken to cover a distance by an object with speed is given by the formula;

Time = distance/speed

Substituting into the equation, we have;

408/x = 405/x + 6

Cross-multiplying, we have;

408*(x + 6) = 405*x

408x + 2448 = 405x

408x - 405x = 2448

3x = 2448

x = 2448/3

x = 816 kilometers per hour.

3 0
3 years ago
Two pairs of statements about Jeff's housing expenditures are given below. How can both pairs of statements be true?
Irina18 [472]

Explanation:

Pair 1 is true if Jeff's monthly income is $600/20% = $3,000.

Pair 2 is true if Jeff's monthly income is $1200/10% = $12,000.

Both pairs can be true if Jeff's monthly income increased by a factor of 4 in the 20 years from 1990 to 2010.

Obviously, Jeff spent more on housing in 2010. (Fortunately for Jeff, that larger expenditure was a smaller fraction of his income.)

5 0
3 years ago
Need help for this please
Ksju [112]

Answer:

The area of the circle is approximately <u>154</u> square meters

Step-by-step explanation:

Area of a circle formula: A = \pi r^{2}

given:

- diameter = 14m

- pi = 22/7

radius (r) = diameter/2 = 14/2 = 7\\A = 22/7*(7)^2\\A = 22/7*(49)\\A = 154

8 0
2 years ago
Prove the function f: R- {1} to R- {1} defined by f(x) = ((x+1)/(x-1))^3 is bijective.
Eduardwww [97]

Answer:

See explaination

Step-by-step explanation:

given f:R-\left \{ 1 \right \}\rightarrow R-\left \{ 1 \right \} defined by f(x)=\left ( \frac{x+1}{x-1} \right )^{3}

let f(x)=f(y)

\left ( \frac{x+1}{x-1} \right )^{3}=\left ( \frac{y+1}{y-1} \right )^{3}

taking cube roots on both sides , we get

\frac{x+1}{x-1} = \frac{y+1}{y-1}

\Rightarrow (x+1)(y-1)=(x-1)(y+1)

\Rightarrow xy-x+y-1=xy+x-y-1

\Rightarrow -x+y=x-y

\Rightarrow x+x=y+y

\Rightarrow 2x=2y

\Rightarrow x=y

Hence f is one - one

let y\in R, such that f(x)=\left ( \frac{x+1}{x-1} \right )^{3}=y

\Rightarrow \frac{x+1}{x-1} =\sqrt[3]{y}

\Rightarrow x+1=\sqrt[3]{y}\left ( x-1 \right )

\Rightarrow x+1=\sqrt[3]{y} x- \sqrt[3]{y}

\Rightarrow \sqrt[3]{y} x-x=1+ \sqrt[3]{y}

\Rightarrow x\left ( \sqrt[3]{y} -1 \right ) =1+ \sqrt[3]{y}

\Rightarrow x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

for every y\in R-\left \{ 1 \right \}\exists x\in R-\left \{ 1 \right \} such that x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

Hence f is onto

since f is both one -one and onto so it is a bijective

8 0
3 years ago
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