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Tanzania [10]
3 years ago
10

bag contains 6 red marbles and 12 yellow marbles. If a representative sample contains 3 red ​marbles, then how many marbles woul

d you expect it to​ contain? Explain.
Mathematics
1 answer:
Dafna11 [192]3 years ago
7 0
I think 6 because it’s a ratio

Because the yellow would always be twice as much as the red so 3x2 is 6 just like 6x2=12
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I think it would be “Literal”.
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The exprnditure of 15 days of a man is as follows. calculate this average expenditure. 30,25,35,24,40,35,30,25,
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Answer:

$30.53

Step-by-step explanation:

Find total sum

30+25+35=90

90+24=114

114+40=154

154+35=189

…

You will get 458

458/15=30.53333

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Plzzzzzzzz help for a brainly
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Answer:

The first one is supplementary

The second one is complementary

The third one is vertical

Step-by-step explanation:

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Aaron is standing at point C, watching his friends on a Ferris wheel. He knows that he is looking up at a 57° angle and the meas
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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 seco
kari74 [83]

Answer:

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.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

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

total number of possible cases =36

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}}

p(\text{A})=\frac{5}{36}

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}

Now,

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

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

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.

3 0
3 years ago
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