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seraphim [82]
3 years ago
9

The length of a rectangle is the width minus 8 units. The area of the rectangle is 9 units. What is the length, in units, of the

rectangle?
Mathematics
1 answer:
frez [133]3 years ago
5 0

Length of rectangle is 1 unit  

<u>Solution: </u>

Given that  

Length of a rectangle is width minus 8 units.

Area of rectangle = 9 square units

Need to calculate the length of rectangle.

Let us assume width of rectangle = x

As Length is width minus 8,  

Length of the rectangle = x – 8

\text{ Area of rectangle }=\text{ width of rectangle }\times \text{ Length of the rectangle }

\text{ Area of rectangle }= x\times (x-8)

\text { Area of rectangle }=\left(x^{2}-8 x\right)

As given that area of rectangle = 9 square units

\begin{array}{l}{\Rightarrow x^{2}-8 x=9} \\\\ {\Rightarrow x^{2}-8 x-9=0}\end{array}

On solving above quadratic equation for x, we get

\begin{array}{l}{\Rightarrow x^{2}-9 x+x-9=0} \\\\ {\Rightarrow x(x-9)+1(x-9)=0} \\\\ {\Rightarrow (x-9)(x+1)=0}\end{array}

When x-9 =9, x = 9

When x + 1 = 0, x = -1

As width cannot be negative, so considering x = 9  

\text{ Length of rectangle }= x-8 = 9-8 = 1 \text{ unit }

Hence, Length of rectangle is 1 unit.

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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


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4 years ago
66 points! for if you get this right 99 kids go race go karts for a all day pass is $56 per kid then they play arcade games at a
Romashka-Z-Leto [24]

The total amount that was spent is $6099

<h3>How to determine the total amount that was spent?</h3>

The given parameters are

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So, we have

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Evaluate the expression

Total = 6099

Hence, the total amount that was spent is $6099

Read more about expressions at

brainly.com/question/723406

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A. 1 /14

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3 1/2 / 1/2 = 1 /14

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