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antoniya [11.8K]
2 years ago
9

What is the solution of the system

Mathematics
1 answer:
ozzi2 years ago
7 0

The solution to the system of equation is (14, 6)

1 / 2 x = 10 - 1 / 2 y

8 = x -y

Therefore, let's combine the equation.

<h3>Simultaneous equation:</h3>

1 / 2 x + 1 / 2 y = 10

x - y = 8

x + y = 20

x - y = 8

2y = 12

y = 12 / 2

y = 6

x = 20 - 6

x = 14

Therefore, the solution of the system of equation is (14, 6)

learn more on system of equation here: brainly.com/question/8914228?referrer=searchResults

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-5*-2=-10 The factors to multiply to give negative 10 is -5 and 2 6 plus negative two gives you negative four. The factors to add to give you -4 is 6 and -2
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Step-by-step explanation:

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PLZ HELP ME! Whoever gets it right will be marked brainiest
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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

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

         =0.784

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