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2 years ago

9

Please find the median, mode, range, and the Mean Absolute Deviation for the following

1 answer:

Ronch [10]2 years ago

7 0

Answer:

The median is the first number the mode is middle number and rage is last number

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jet skis rent for $25 per hour plus a flat fee of $50. Noel can rent the jet ski for 6 hours. what is the largest value in the r

telo118 [61]

Answer:

The largest value in this situation would be $25+$50x6 which equals $450

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2 years ago

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The difference of d minus 3 divided by 2?

Allisa [31]

$\text{Hey there!}$

$\text{What is the difference of d minus 3 divided by 2?}\\\\\text{Difference means subtract/subtraction}\\\\\text{Minus would also be subtraction}\\\\\text{Divided by is division}\\\\\text{d would be the start of the problem}\\\\\text{the 3 divided by 2 would be on the right side ( the second part of the equation)}\\\\\boxed{\boxed{\bf{\ Answer:: d-\dfrac{3}2{}}}}}\checkmark$

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2 years ago

joe drove 408 miles using 16 gallons of gas at this ratr how many mokrs would he drive using 11 gallons of gas

stiv31 [10]

I was on the right track, lol. 280.5 miles.

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2 years ago

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In both of the problems where would you put parentheses to make it statement true?

Darina [25.2K]

In the first one it would be (21-8)-6=7.
and in the second one it would be 21-(8-6)=19

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2 years ago

What’s is the answer

kvasek [131]

When you are multiplying an exponent directly into a number/variable with an exponent, you multiply the exponents together.

For example:

$(x^{2} )^{3} = x^6$

$(x^{3} )^5=x^{15}$

When you are multiplying a variable with an exponent by another variable with an exponent, you add the exponents together.

For example:

$(x^{2} )(x^{3})=x^{5}$

$(x^{1} )(x^{2})=x^{3}$

$(\frac{(x^{-3})(y^{2})}{(x^{4})(y^{6})} )^{3}=\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}$

You multiply 3 into each exponent in the numerator and the denominator

$\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}= \frac{y^{6}}{(x^{9})(x^{12})(y^{18})}$

When you have a negative exponent, you move it to the other side of the fraction to make the exponent positive.

$\frac{y^{6}}{(x^{21})(y^{18})} = \frac{1}{(x^{21})(y^{12})}$

When you have something like this:

$\frac{x^{2}}{x^5}$

You subtract the exponents together, so:

$\frac{x^2}{x^5} = x^{2-5} = x^{-3} = \frac{1}{x^3}$

Your answer is the second option

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2 years ago