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2 years ago

Can some smart person help me out here!!!

Mathematics

1 answer:

vitfil [10]2 years ago

8 0

Answer:

the answer is C I'm studying this right now

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An experiment was conducted to observe the effect of an increase in temperature on the potency of an antibiotic. Three 1-ounce p

ludmilkaskok [199]

Answer:

a) $y=-0.317 x +46.02$

b) Figure attached

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

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.

a) Find the least-squares line appropriate for this data.

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:

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

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

And we can find the intercept using this:

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

So the line would be given by:

$y=-0.317 x +46.02$

b) Plot the points and graph the line as a check on your calculations.

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

c) Calculate S^2

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:

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

We have that SST= SSR+SSE, and then SSE=SST-SSR= 792-601.67=190.33[/tex]

So then :

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

5 0

3 years ago

20 points!!!

UkoKoshka [18]

150.

You can read the uppermost point on the graph which is pretty much that point.

4 0

3 years ago

Read 2 more answers

A triangle with side lengths 13, 30, and 42 is a right triangle. True or False?

kaheart [24]

Answer:

False

Step-by-step explanation:

$42^2-30^2-13^2=695$

695 ≠ 0

4 0

3 years ago

A cake manufacturer boxes Swiss chocolate and German chocolate cakes at one site. A box of Swiss chocolate cake contains 0.55 lb

Kobotan [32]

Answer:

2,500 German chocolate cake boxes.

1,500 Swiss chocolate cake boxes.

Step-by-step explanation:

Let 'S' be the number of Swiss chocolate cakes boxed and 'G' the number of German cholocate cakes boxed. If all of the available ingredients are used:

$0.55*S +0.59*G = 2,300\\0.45*S+0.41*G = 1,700$

Solving the linear system above:

$(0.55 *S +0.59*G)- \frac{0.55}{0.45} (0.45*S+0.41*G)= 2,300-\frac{0.55}{0.45}*1,700*\\0.088888*G = 222.222\\G=2,500\\S =\frac{2,300-2,500*0.59}{0.55}\\S=1500$

2,500 German chocolate cake boxes and 1,500 Swiss chocolate cake boxes can be made each day.

6 0

3 years ago

I need some help! please help me if you know the answers! will give brainliest!

miv72 [106K]

C because you have to round and it would be C

7 0

3 years ago

Read 2 more answers