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Andrews [41]
2 years ago
5

8.

Mathematics
1 answer:
mixer [17]2 years ago
7 0

Answer:

{2, 3, 4, 6, 8, 9, 12, 18}

Step-by-step explanation:

If A= {x: xis a factor of 72}

A = {2, 3, 4, 6, 8, 9, 12, 18, 36, 72}

If B= {x: 2x+3<2x-3).

-2x+3 <2x-3

-2x - 2x < -3-3

-4x < -6

x > 3/2

If

C= {x:x<20}

C = {-∞, 19}

The intersection of the sets AnBnc = {2, 3, 4, 6, 8, 9, 12, 18}

Note that intersection means all the elements common to the three sets

<em></em>

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Answer:

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

where A is the event of getting a prize on opening a cereal box

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:

  • Case 1, Where Hannah gets prize when she buys the first box:

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

Let N be the event of Hannah winning the prize on buying 3 or less boxes before getting a prize

⇒N=K∪L∪M

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