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Temka [501]
3 years ago
8

If you invest money, after a certain period of time you’ll get interest. For example, if you invest $2,000 at 5% annual interest

rate, in a year you will have $2,000∙0.05=$100 of annual simple interest in addition to the original amount. If you invest $250.00 at 4% interest rate, what will be the interest?
Mathematics
2 answers:
Rashid [163]3 years ago
7 0

Answer:

The interest at 4 % of $250.00 will be $10

Step-by-step explanation:

If you invest money, after a certain period of time you’ll get interest.

For example, if you invest $2,000 at 5 % annual interest rate, in a year you will have ( $2,000 ) x ( 0.05 ) = $100 of annual simple interest in addition to the original amount.

If you invest $250.00 at 4% interest rate, the interest will be ( $250.00 ) x( 0.04 ) = $10.

Simple interest is based on arithmetic progression and compound interest is based on geometric progression.

Gekata [30.6K]3 years ago
6 0
$250,000 x 0.04 = 10,000 in addition to the original amount
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Answer: C.) $51

Step-by-step explanation:

$85.00 x 40%

85.00 x 0.40= 34

85.00-34=51.00

$51

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gtnhenbr [62]

Answer:

a) r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857  

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

b) m=\frac{64.5}{2000}=0.03225  

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

\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50  

\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425  

b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27  

So the line would be given by:  

y=0.3225 x -0.27  

Step-by-step explanation:

Part a

The correlation coeffcient is given by this formula:

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=4 \sum x = 200, \sum y = 5.37, \sum xy = 333, \sum x^2 =12000, \sum y^2 =9.3501  

r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857  

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

Part b

m=\frac{S_{xy}}{S_{xx}}  

Where:  

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}=12000-\frac{200^2}{4}=2000  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=333-\frac{200*5.37}{4}=64.5  

And the slope would be:  

m=\frac{64.5}{2000}=0.03225  

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

\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50  

\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425  

And we can find the intercept using this:  

b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27  

So the line would be given by:  

y=0.3225 x -0.27  

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3 years ago
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Aleonysh [2.5K]
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a baseball coach collected data to analyze the free throw shooting percentages of players during a game and during practice. The
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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