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S_A_V [24]
3 years ago
13

Find the MAD:58, 38, 54, 48, 26, 36​

Mathematics
2 answers:
trasher [3.6K]3 years ago
6 0

Answer:

10.166666666667

Step-by-step explanation:

Lemur [1.5K]3 years ago
3 0

Answer:

(add up all the number and divide it by 6, cause there are about 6 numbers in the list)

58 + 38 + 54 + 48 + 26 + 36 = 260 ÷ 6 = 43.3

(subtract the mean {43.3} with the same listed numbers)

43.3 - 58 = 14.7

43.3 - 38 = 5.3

43.3 - 54 = 10.7

43.3 - 48 = 4.7

43.3 - 26 = 17.3

43.3 - 36 = 7.3

(add up all the difference and divide it with 6 again)

14.7 + 5.3 + 10.7 + 4.7 + 17.3 + 7.3 = 60 ÷ 6 = 10

so our MAD (mean absolute deviation) would be 10!

~hope this helps, have a good day/afternoon/nigh~

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C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

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Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

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Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

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Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

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probability that sum of cards 8 is and one of cards is 5

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p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

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