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True [87]
2 years ago
6

Can someone answers this question please master it correctly if it’s corect I will mark you brainliest

Mathematics
1 answer:
Agata [3.3K]2 years ago
6 0

Answer:

15 daffodil bouquets

Step-by-step explanation:

Out of 9 bouquets that are sold, 3 were daffodil bouquets

3/9 = simplify as 1/3

1/3 = x/45              Cross multiply

3x = 45                  Divide 3 on both sides

x = 15

<h3><u><em>Hope this helps!!!</em></u></h3><h3><u><em>\ :) /</em></u></h3>
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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
3 years ago
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