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3 years ago

Which statistical measure changes when every number in a data set is increased by 10

Mathematics

2 answers:

jarptica [38.1K]3 years ago

8 0

Answer:

There are various statistical measures that can be affected by change of origin, they include; the mode, the median, the mean

Step-by-step explanation:

The variance and standard deviation remain unchanged

timama [110]3 years ago

7 0

Answer:

All will change except, range, IQR, variance and std deviation

Step-by-step explanation:

Given a data set we can find descriptive statistics such as mean, median, mode, std deviation, variance, max, minimum etc.

If every number in a data set is increased by 10 we find

that variance and std deviation will not change.

This is because when mean varies by +10 and each entry by +10 variance remains the same and hence std deviation remains the same.

Next is the interquartile range, which is the difference between I quartile and II quartile, this will remain the same

Similarly, range the difference between minimum and maximum will remain the same.

Except these four all others, namely, mean, median, mode, min, maximum , I quartile and III quartile will all increase by 10.

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Read 2 more answers

Need help with my homework

Volgvan

Answer:

$\displaystyle y=\frac{16-9x^3}{2x^3 - 3}$

$\displaystyle y=-\frac{56}{13}$

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

$\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}$

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

$\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3$

Simplify the radical with the cube:

$\displaystyle x^3=\frac{3y+16}{2y+9}$

Multiply by 2y+9

$\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)$

Simplify:

$\displaystyle x^3(2y+9)=3y+16$

Operate the parentheses:

$\displaystyle x^3(2y)+x^3(9)=3y+16$

$\displaystyle 2x^3y+9x^3=3y+16$

Subtract 3y and $9x^3$ :

$\displaystyle 2x^3y - 3y=16-9x^3$

Factor y out of the left side:

$\displaystyle y(2x^3 - 3)=16-9x^3$

Divide by $2x^3 - 3$ :

$\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}$

ii) To find y when x=2, substitute:

$\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}$

$\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}$

$\displaystyle y=\frac{16-72}{16- 3}$

$\displaystyle y=\frac{-56}{13}$

$\mathbf{\displaystyle y=-\frac{56}{13}}$

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