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sergejj [24]
3 years ago
9

Rin incorrectly calculated the mean absolute deviation for the data set 25,25,31,36,38.Her work is shown here. 1. Find the mean.

Mean = 25 + 25 + 31 + 36 + 38 5 = 155 5 = 31 2. Find each absolute deviation. Absolute deviations:6,6,0,5,7 3. Find the mean absolute deviation. MAD = 6 + 0 + 5 + 7 4 = 18 4 = 4.5 In which step did Rin make an error?
Mathematics
2 answers:
dexar [7]3 years ago
3 0
In Step 3 was Her Mistake
lara31 [8.8K]3 years ago
3 0

Answer:

Rin did mistake in 3rd step

Step-by-step explanation:

Mean is correct  according to rin's calculation

Absolute mean deviation is  |data value-mean|

Hence, values of absolute deviations are also correct

To find the Mean absolute deviation we sum the differences of absolute deviations and then divide by the number of observations

it should be 6+6+0+7+5=24/5=4.8

Not 18/4

Hence, Rin did mistake in 3rd step

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Answer:

Step-by-step explanation:

x = ¼y

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A box of candy hearts contains 52 hearts, of which 19 are white, 10 are tan, 7 are pink, 3 are purple, 5 are yellow, 2 are orang
laiz [17]

<u>Answer-</u>

a. Probability that  three of the candies are white = 0.29

b. Probability that three are white, 2 are tan, 1 is pink, 1 is yellow, and 2 are green = 0.006

<u>Solution-</u>

There are 19 white candies, out off which we have to choose 3.

The number of ways we can do the same process =

\binom{19}{3} = \frac{19!}{3!16!} = 969

As we have to draw total of 9 candies, after 3 white candies we left with 9-3 = 6, candies. And those 6 candies have to be selected from 52-19 = 33 candies, (as we are drawing candies other than white, so it is subtracted)

And this process can be done in,

\binom{33}{6} = \frac{33!}{6!27!} =1107568

So total number of selection = (969)×(1107568) = 1073233392

Drawing 9 candies out of 52 candies,

\binom{52}{9} = \frac{52!}{9!43!} = 3679075400

∴P(3 white candies) = \frac{1073233392}{3679075400} =0.29



Total number of ways of selecting 3 whites, 2 are tans, 1 is pink, 1 is yellow, and 2 are greens is,

\binom{19}{3} \binom{10}{2} \binom{7}{1} \binom{5}{1} \binom{6}{2}

=(\frac{19!}{3!16!}) (\frac{10!}{2!8!}) (\frac{7!}{1!6}) (\frac{5!}{1!4!}) (\frac{6!}{2!4!})

=(969)(45)(7)(5)(15)=22892625

Total number of selection = 3 whites + 2 are tans + 1 is pink + 1 is yellow + 2 greens = 9 candies out of 52 candies is,

\binom{52}{9}=\frac{52!}{9!43!} =3679075400

∴ P( 3 whites, 2 are tans, 1 is pink, 1 is yellow, 2 greens) =

\frac{22892625}{3679075400} = 0.006


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