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skad [1K]
3 years ago
8

Suppose that 3 is a factor of a, a is a divisor of 12, and a is positive. What is the number of possible values of a?

Mathematics
2 answers:
klio [65]3 years ago
8 0

Answer:

3

Step-by-step explanation:

12 is divisible by: 1, 2, 3, 4, 6, 12

Because 3 is a factor of a, that means 3 multiplied by another number equals a. Additionally, a is a divisor of 12, meaning a multiplied by another number equals 12.

Out of the numbers, 12 is divisible by, only 3 are also divisible by 3: 3, 6, 12 This gives three possible values of a.

Zina [86]3 years ago
5 0
Answer: the number of possible values of a are 3 and 6

Explanation:
Since a is a divisor of 12
Then let’s see what are the divisor for 12
12= 6 x 2
12= 3 x 4
And we know that 3 is a factor of a
So if a was 3 then 3/3 is exactly 1
And if a was 6 then 6/3 is exactly 2 with no remainder

And 3 and 6 are also the divisor of 12
12/6=2
12/3=4
So a should be 3 and 6

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Probability that Hannah only has to buy 3 or less boxes before getting a prize is 0.784

Step-by-step explanation:

Given, 40% of cereal boxes contain a prize

⇒probability of getting a prize on opening a box, P(A)=0.4

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and probability of not getting a prize on opening a box, P(A')=1-P(A)=0.6

where A' is the event of not getting a prize on opening a cereal box

This problem needs to be divided into 3 situation:

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Let K be the event of Hannah winning the prize on buying the first box.

⇒P(K)=P(A)=0.4

  • Case 2, Where Hannah gets prize when she buys the second box:

I<u>n this event Hannah should not get the prize in first box but should get the prize on buying the second box</u>

Let L be the event of Hannah winning the prize on buying the second box

So, P(L)=P(A')·P(A)

           =(0.6)·(0.4)

           =0.24

  • Case 3,Where Hannah gets prize when she buys the third box:

<u>In this event Hannah should not get the prize in first and second box but should get the prize on buying the third box</u>

Let L be the event of Hannah winning the prize on buying the third box

So, P(L)=P(A')·P(A')·P(A)

           =(0.6)·(0.6)·(0.4)

           =0.144

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⇒N=K∪L∪M

Now, Required probability is P(N)=P(K∪L∪M)=P(K)+P(L)+P(M) [As events K,L and M are independent and disjoint events]

⇒P(N)=0.4+0.24+0.144

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