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Tems11 [23]
1 year ago
11

Claudia invested $12,000 at an interest rate of 4.5% compounded monthly. In the formula A=p(1+r/n)^nt, A is the worth of an acco

unt after t years, n is the number of times that interest is compounded, P is the initial deposit, and r is the interest rate.Approximately how much money will be in Claudia’s account after 7 years?
Mathematics
1 answer:
aleksandrvk [35]1 year ago
6 0

Answer: $16,433.42

Step-by-step explanation:

Hi, to answer this question we have to apply the formula given:

A=p (1+r/n)^nt,

  • r must be in decimal form (percentage divided by 100)
  • n is equal to 12 ( it compounds 12 times per year)

Replacing with the values given:

A = 12,000 (1+ 0.045/12) ^12(7)

Solving:

A = 12,000 (1+ 0.045/12) ^12(7)

A= 12,000 (1.00375) ^84

A = $16,433.42

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Ethan bought 4 packages of pencils. After he gave 8 pencils to his friends, he had 40 pencils left over. How many pencils were i
swat32

Answer:

12 pencils in each package.

Step-by-step explanation:

Given:

Ethan bought 4 packages of pencils.

No of Packages = 4

No of pencils given to friend = 8 pencils

No of pencils left over =40 pencils

∴Total Number of Pencils = No of pencils given to friend + No of pencils left over =40+8 =48

Total Number of pencils in each package = \frac{\textrm{Total Number of Pencils}}{\textrm{No Of Packages}}= \frac{48}{4}= 12 \ pencils

8 0
2 years ago
What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?
astraxan [27]
ANSWER


The value of the expression is
- 1


EXPLANATION

Method 1: Rewrite as product of
{i}^{2}


The expression given to us is,

{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}


We use the fact that
{i}^{2}  =  - 1
to simplify the above expression.



{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  {i}^{0}  \times {i}^{1}  \times {i}^{3}   \times {i}^{2}   \times {i}^{4}


This implies,


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  {i}^{0}  \times {i}^{2}  \times {i}^{2}   \times {i}^{2}   \times {i}^{2} \times {i}^{2}


We substitute to obtain,

{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  1\times  - 1 \times  - 1  \times  - 1\times  - 1 \times  - 1


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  1\times  1 \times   1  \times  - 1 =  - 1


Method 2: Use indices to solve.



{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  = {i}^{0 + 1 + 2 + 3 + 4}



This implies that,


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  = {i}^{10}




{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  (  {{i}^{2}} )^{5}


{i}^{0}  \times {i}^{1}  \times {i}^{2}  \times {i}^{3}  \times {i}^{4}  =  (   - 1 )^{5}   =  - 1


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2 years ago
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8 total pens....4 are black

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