What numbers can give me 8 when added and 30 when multiplied

1 answer:

7 0

In this problem, it is important to take note that the number of numbers to be utilized isn't specified so it can be up to a thousand numbers. It wasn't also specified if repeating of numbers is allowed or not. So with those taken into consideration and the condition presented in mind, the numbers that can give you 8 when added and 30 when multiplied are 2, 3, 5, -1, and another -1. The derivation from this is mainly from factorization and a little bit of logic.

here is the solution.

2 x 3 x 5 x -1 x -1 = 30
6 x 5 x -1 x -1 = 30
30 x -1 x -1 = 30
-30 x -1 = 30
30 = 30

2 + 3 + 5 + -1 + -1 = 8
5 + 5 + -1 + -1 = 8
10 + -1 + -1 = 8
9 + -1 = 8
8 = 8

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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

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