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likoan [24]
3 years ago
11

3x+8y=27 4x+3y=13 solve the simultaneous equation

Mathematics
1 answer:
dmitriy555 [2]3 years ago
4 0

Step-by-step explanation:

3x+8y=27

3x=27-8y

x=27-8y/3

4(27-8y/3) + 3y=13

36-32/3y +3y =13

36-23/3y=13

36-13=23/3y

y=3

x=1

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3 years ago
One urn contains one blue ball (labeled B1) and three red balls (labeled R1, R2, and R3). A second urn contains two red balls (R
marusya05 [52]

Answer:

(a) See attachment for tree diagram

(b) 24 possible outcomes

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Given

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If urn 1 is selected, the following selection exists:

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

If urn 2 is selected, the following selection exists:

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<em>See attachment for possibility tree</em>

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<u>For urn 1</u>

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n = \{B_1,R_1,R_2,R_3\}

Each of the balls has 3 subsets. i.e.

B_1 \to [R_1, R_2, R_3]; R_1 \to [B_1, R_2, R_3]; R_2 \to [B_1, R_1, R_3]; R_3 \to [B_1, R_1, R_2]

So, the selection is:

Urn\ 1 = 4 * 3

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<u>For urn 2</u>

There are 4 balls in urn 2

n = \{B_2,B_3,R_4,R_5\}

Each of the balls has 3 subsets. i.e.

B_2 \to [B_3, R_4, R_5]; B_3 \to [B_2, R_4, R_5]; R_4 \to [B_2, B_3, R_5]; R_5 \to [B_2, B_3, R_4]

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Total number of outcomes is:

Total = Urn\ 1 + Urn\ 2

Total = 12 + 12

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