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adell [148]
3 years ago
11

You begin with $90 in your savings account and your friend begins

Mathematics
2 answers:
Colt1911 [192]3 years ago
7 0

Answer:

F(f) = 15t + 35 represents the total amount of savings your friend would make in t weeks.

F(d) = 10t + 90 represents the total amount of saving you, darian, would make in t weeks.

When you graph the equations, plugging in different values for t, you can see that the graphs  intersect at (11,200). This means that at 11 weeks, both you and your friend have the same amount of money saved up, $200.  They will not have the same amount of money in 10 weeks.

BlackZzzverrR [31]3 years ago
6 0

Answer:

<h3>a. y=10x+90</h3>

y=15x+35

<h3>b. Your friend is wrong.</h3>

Step-by-step explanation:

If you begin with $90 and add $10 each week, then we express it as

y=10x+90

Where x refers to weeks and y refers to the total amount of money in you savings account.

On the other hand, your friend begins with $35 and deposits $15 each week, the equation that expresses your friend's savings is

y=15x+35

Therefore, the sytem of linear equation that models this situation is

y=10x+90

y=15x+35

The graph attached shows this system. There you can notice that the solution is (11, 200), which means in 11 weeks both will have the same amount saved.

Therefore, your friend is wrong, because after 10 weeks you won't have the same amount.

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a) y=-0.317 x +46.02

b) Figure attached

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Step-by-step explanation:

We assume that th data is this one:

x: 30, 30, 30, 50, 50, 50, 70,70, 70,90,90,90

y: 38, 43, 29, 32, 26, 33, 19, 27, 23, 14, 19, 21.

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For this case we need to calculate the slope with the following formula:

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}

So we can find the sums like this:

\sum_{i=1}^n x_i = 30+30+30+50+50+50+70+70+70+90+90+90=720

\sum_{i=1}^n y_i =38+43+29+32+26+33+19+27+23+14+19+21=324

\sum_{i=1}^n x^2_i =30^2+30^2+30^2+50^2+50^2+50^2+70^2+70^2+70^2+90^2+90^2+90^2=49200

\sum_{i=1}^n y^2_i =38^2+43^2+29^2+32^2+26^2+33^2+19^2+27^2+23^2+14^2+19^2+21^2=9540

\sum_{i=1}^n x_i y_i =30*38+30*43+30*29+50*32+50*26+50*33+70*19+70*27+70*23+90*14+90*19+90*21=17540

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}=49200-\frac{720^2}{12}=6000

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=17540-\frac{720*324}{12}{12}=-1900

And the slope would be:

m=-\frac{1900}{6000}=-0.317

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

\bar x= \frac{\sum x_i}{n}=\frac{720}{12}=60

\bar y= \frac{\sum y_i}{n}=\frac{324}{12}=27

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b=\bar y -m \bar x=27-(-0.317*60)=46.02

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For this case we can use excel and we got the figure attached as the result.

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In oder to calculate S^2 we need to calculate the MSE, or the mean square error. And is given by this formula:

MSE=\frac{SSE}{df_{E}}

The degred of freedom for the error are given by:

df_{E}=n-2=12-2=10

We can calculate:

S_{y}=\sum_{i=1}^n y^2_i -\frac{(\sum_{i=1}^n y_i)^2}{n}=9540-\frac{324^2}{12}=792

And now we can calculate the sum of squares for the regression given by:

SSR=\frac{S^2_{xy}}{S_{xx}}=\frac{(-1900)^2}{6000}=601.67

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S^2=\hat \sigma^2=MSE=\frac{190.33}{10}=19.03

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