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Serggg [28]
3 years ago
10

What is the solution to the following system? [x+y+z=6 x-y+Z=8 x+y-2-0

Mathematics
2 answers:
Alex Ar [27]3 years ago
7 0

x+y+z=6

x-y+z=8

x+y-2=0   (fixed typo)

Subtracting the first two,

2y = -2

y = -1

x = -y+2 = 3

z=6-x-y=4

Answer: (3, -1, 4), first choice

Inessa [10]3 years ago
6 0

Answer:

144

Step-by-step explanation:

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Step-by-step explanation:

yuson has to finish am total of 30 hours in community service

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in x days, hour of service done=2x

after x days hours of service left=30-2x

therefore the linear equation representing the hours yuson has left after x days is:

30-2x

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Step-by-step explanation:

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PLZ HELP ME! Whoever gets it right will be marked brainiest
Brut [27]

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

7 0
3 years ago
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