How do you find the domain and range of $\sqrt{x-8}$?

Parker Bird
2022-07-20
Answered

How do you find the domain and range of $\sqrt{x-8}$?

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Killaninl2

Answered 2022-07-21
Author has **20** answers

The square root is real only when the radicand is positive, or at least equal to zero. So the domain is going to be whenever

$x-8\ge 0$

$x\ge 8$

Using interval notation, we say the domain of x is $[8,\mathrm{\infty})$

The range is all the the y values that result from this domain. So the range starts at x=8

$y=\sqrt{8-8}=\sqrt{0}=0$

and goes up to infinity. Using interval notation the range of y is $[0,\mathrm{\infty})$

You can also see this by inspection in the graph

graph{$\sqrt{x-8}[-1,17,-2,5]$}

$x-8\ge 0$

$x\ge 8$

Using interval notation, we say the domain of x is $[8,\mathrm{\infty})$

The range is all the the y values that result from this domain. So the range starts at x=8

$y=\sqrt{8-8}=\sqrt{0}=0$

and goes up to infinity. Using interval notation the range of y is $[0,\mathrm{\infty})$

You can also see this by inspection in the graph

graph{$\sqrt{x-8}[-1,17,-2,5]$}

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Most z-score table stops at 3.4. If this is right, kindly explain how it was gotten. Thanks

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Trying to find the probability $b/w$ two scores. I know mapping to $z$-score is:

${z}_{11}=(11-10)/1.5=.67$

and

${z}_{14}=(14-10)/1.5=2.67$

So,

$\overline{x}=10,\text{}\sigma =\mathrm{1.5.}$

Would I simple subtract the two from each other? and so on the real number line it would range from 0 to 100 and that is how the distribution would be set up. So, 2.67−2=2?

On the z-table 2 is given a value of 0.9772.

My teacher likes to give things as either a proportion or a percentage so how might one be able to address the two. Like if this question was asking for the proportion of scores that fall between 11 and 14 or if the question said what's the percentage of scores that fall between 11 and 14. How might be able to address both scenarios?

Just confused on the ideas of proportion, probability, and percentage and how they are relate.

${z}_{11}=(11-10)/1.5=.67$

and

${z}_{14}=(14-10)/1.5=2.67$

So,

$\overline{x}=10,\text{}\sigma =\mathrm{1.5.}$

Would I simple subtract the two from each other? and so on the real number line it would range from 0 to 100 and that is how the distribution would be set up. So, 2.67−2=2?

On the z-table 2 is given a value of 0.9772.

My teacher likes to give things as either a proportion or a percentage so how might one be able to address the two. Like if this question was asking for the proportion of scores that fall between 11 and 14 or if the question said what's the percentage of scores that fall between 11 and 14. How might be able to address both scenarios?

Just confused on the ideas of proportion, probability, and percentage and how they are relate.

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Standardized effect-sizes are reported in the same units as the original measures.

Standardized effect-sizes are very generalizable, allowing results to be pooled across studies.

Standardized effect-sizes are not useful in statistical calculations (e.g., power).

Standardized effect-sizes are reported in the same units as the original measures.

Standardized effect-sizes are very generalizable, allowing results to be pooled across studies.