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NikAS [45]
3 years ago
6

Suppose you take $25,000 and deposit some of it into a savings account that pays 2.2% interest. You then take the rest of the mo

ney and deposit it in a CD that pays 3.05% interest at the end of 10 years the value of both accounts is $31,648. How much was deposited into each account?
Mathematics
1 answer:
FromTheMoon [43]3 years ago
3 0

Answer:

The amount deposited in the savings account is $11,494.12

The amount deposited in the CD account is $13,505.88

Step-by-step explanation:

Firstly, let’s calculate the amount of interest;

Interest = Amount at end of 10 years - amount deposited= 31,648 -25,000 = 6,648

Let the amount deposited in the savings account be $x while the amount deposited in the CD account be $y

Mathematically;

x + y = 25,000 ••••••••••(i)

Now the formula for simple interest is;

PRT/100

where P is the amount deposited, r is the rate and T is the time

On the savings account, amount of interest generated will be;

(x * 2.2 * 10)/100 = 22x/100

On the CD act, amount of interest generated will be;

(y * 3.05 * 10)/100 = 30.5y/100

The addition of both give the total interest;

22x/100 + 30.5y/100 = 6648

Multiply through by 100

22x + 30.5y = 664800 •••••••(ii)

From i, x = 25,000 - y

Substitute this into ii

22(25000-y) + 30.5y = 664800

550,000 -22y + 30.5y = 664,800

8.5y = 664,800-550,000

8.5y = 114,800

y = 114,800/8.5

y = $13,505.88

x = 25,000 -y

x = 25,000 - 13,505.88

x = $11,494.12

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Answer:

a) y=0.00991 x +1.042  

b) r^2 = 0.7503^2 = 0.563

c) r=\frac{7(30095)-(4210)(49)}{\sqrt{[7(2595100) -(4210)^2][7(354) -(49)^2]}}=0.7503  

Step-by-step explanation:

Data given

x: 500, 700, 750, 590 , 540, 650, 480

y: 7.00, 7.50 , 9.00, 6.5, 7.50 , 7.0, 4.50

Part a

We want to create a linear model like this :

y = mx +b

Wehre

m=\frac{S_{xy}}{S_{xx}}  

And:  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}  

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}  

With these we can find the sums:  

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=2595100-\frac{4210^2}{7}=63085.714  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=30095-\frac{4210*49}{7}=625  

And the slope would be:  

m=\frac{625}{63085.714}=0.00991  

Nowe we can find the means for x and y like this:  

\bar x= \frac{\sum x_i}{n}=\frac{4210}{7}=601.429  

\bar y= \frac{\sum y_i}{n}=\frac{49}{7}=7  

And we can find the intercept using this:  

b=\bar y -m \bar x=7-(0.00991*601.429)=1.042  

And the line would be:

y=0.00991 x +1.042  

Part b

The correlation coefficient is given by:

r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}  

For our case we have this:

n=7 \sum x = 4210, \sum y = 49, \sum xy = 30095, \sum x^2 =2595100, \sum y^2 =354  

r=\frac{7(30095)-(4210)(49)}{\sqrt{[7(2595100) -(4210)^2][7(354) -(49)^2]}}=0.7503  

The determination coefficient is given by:

r^2 = 0.7503^2 = 0.563

Part c

r=\frac{7(30095)-(4210)(49)}{\sqrt{[7(2595100) -(4210)^2][7(354) -(49)^2]}}=0.7503  

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