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3 years ago

Find the range of the following data set. 1 1/4,5/8,3/4 ,1/2 ,1 1/2, 1 3/4

2 answers:

5 0

1 1/4 = 1 2/8
5/8 = 5/8
3/4 = 6/8
1/2 = 4/8
1 1/2 = 1 4/8
1 3/4 = 1 6/8
So, the biggest number (1 6/8) - the smallest number (4/8) = 10/8
But to subtract fractions, you must do a little rule
1x8=8+6=14/8 - 4/8 which is 10/8 or 1 1/4
The answer is C

Alex73 [517]3 years ago

3 0

Answer: The range of the data is (C) $1\dfrac{1}{4}.$

Step-by-step explanation: We are given to find the range of the following data:

$1\dfrac{1}{4},~\dfrac{5}{8},~\dfrac{3}{4},~\dfrac{1}{2},~1\dfrac{1}{2},~1\dfrac{3}{4}\\\\\\=\dfrac{5}{4},~\dfrac{5}{8},~\dfrac{3}{4},~\dfrac{1}{2},~\dfrac{3}{2},~\dfrac{7}{4}.$

RANGE : The range of a data is given by the difference between the larges and the smallest value in the data.

Arranging the given data in ascending order, we have

$\dfrac{1}{2},~\dfrac{5}{8},~\dfrac{3}{4},~\dfrac{5}{4},~\dfrac{3}{2},~\dfrac{7}{4}.$

So, the largest value and the smallest values of the data are

$L=\dfrac{7}{4},\\\\S=\dfrac{1}{2}.$

Therefore, the range of the data is given by

$R=L-S=\dfrac{7}{4}-\dfrac{1}{2}=\dfrac{7-2}{4}=\dfrac{5}{4}=1\dfrac{1}{4}.$

Thus, option (C) is the correct option.

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Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

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Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

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$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

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