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3 years ago

Does the MAD represent a large or small amount of variability? Why?

Mathematics

1 answer:

natita [175]3 years ago

8 0

Answer:

The mean absolute value deviation is a measure of dispersion which gives the average variation of the data from the mean. In order words, it is the average of the positive distances of each point from the mean. The larger the Mean Absolute Deviation, the greater variability there is in the data (the data is more spread out).

The MAD helps determine whether the set's mean is a useful indicator of the values within the set.

The larger the MAD, the less relevant is the mean as an indicator of the values within the set.

Step-by-step explanation:

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S is in between T and U. ST = x + 4 and SU = 2x. TU = 16. Find the value of x.

solniwko [45]

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pochemuha

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3 years ago

Read 2 more answers

Complete the steps in solving 4/9=x/36

Whitepunk [10]

Isolate the x. Note the equal sign. What you do to one side, you do to the other.

Cross multiply

(4/9)(9)(36) = (x/36)(36)(9)

(4)(36) = (x)(9)

Simplify

144 = 9x

Isolate the x. Divide 9 from both sides

144/9 = 9x/9

x = 144/9

x = 16

16 is your answer

hope this helps

3 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

4 years ago