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liq [111]
3 years ago
8

At the apple pan 4 burgers and 3 fries cost 26.50. 5 burgers and 5 fries cost $36.25. what is the cost for each item?

Mathematics
1 answer:
netineya [11]3 years ago
8 0
Fries are $2.50 and burgers are $4.75.

Our system of equations for this situation would be

\left \{ {{4b+3f=26.50} \atop {5b+5f=36.25}} \right.

We want the coefficients of one of the variables to be the same in order to eliminate one of them; we will make the coefficients of f the same.  To do this, multiply the top equation by 5 and the bottom equation by 3:

\left \{ {{5(4b+3f=26.50)} \atop {3(5b+5f=36.25)}} \right. 
\\
\\ \left \{ {{20b+15f=132.50} \atop {15b+15f=108.75}} \right

We subtract the bottom equation:
\left \{ {{20b+15f=132.50} \atop {-(15b+15f=108.75)}} \right. 
\\
\\ 5b = 23.75

Divide both sides by 5:
5b/5 = 23.75/5
b = 4.75

Burgers are $4.75.

Substitute this into the first equation:
4(4.75) + 3f = 26.50
19 + 3f = 26.50

Subtract 19 from both sides:
19 + 3f - 19 = 26.50 - 29
3f = 7.50

Divide both sides by 3:
3f/3 = 7.50/3
f = 2.50
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8 0
3 years ago
The table shows the average annual cost of tuition at 4-year institutions from 2003 to 2010.
nata0808 [166]

Answer: 1) The best estimate for the average cost of tuition at a 4-year institution starting in 2020 =$ 31524.31

2) The slope of regression line b=937.97 represents the rate of change of  average annual cost of tuition at 4-year institutions (y) from 2003 to 2010(x).  Here,average annual cost of tuition at 4-year institutions is dependent on school years .

Step-by-step explanation:

1) For the given situation we need to find linear regression equation Y=a+bX for the given situation.

Let x be the number of years starting with 2003 to 2010.

i.e. n=8

and y be the average annual cost of tuition at 4-year institutions from 2003 to 2010.  

With reference to table we get

\sum x=36\\\sum y=150894\\\sum x^2=204\\\sum xy=718418

By using above values find a and b for Y=a+bX, where b is the slope of regression line.

a=\frac{(\sum y)(\sum x^2)-(\sum x)(\sum xy)}{n(\sum x^2)-(\sum x)^2}=\frac{150894(204)-(36)718418}{8(204)-(36)^2}=\frac{30782376-25863048}{1632-1296}=\frac{4919328}{336}\\\\=14640.85

and

b=\frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}=\frac{8(718418)-(36)150894}{8(204)-(36)^2}=\frac{5747344-5432184}{1632-1296}=\frac{315160}{336}\\\\=937.97


∴ To find average cost of tuition at a 4-year institution starting in 2020.(as n becomes 18 for year 2020 if starts from 2003 ⇒X=18)

So, Y= 14640.85 + 937.97×18 = 31524.31

∴The best estimate for the average cost of tuition at a 4-year institution starting in 2020 = $31524.31


4 0
4 years ago
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