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Feliz [49]
2 years ago
12

I need #3 done ASAP (Must show work) Will give brainliest

Mathematics
1 answer:
Alenkinab [10]2 years ago
4 0

Step-by-step explanation:

<u>Statements                                                  Reasons                                    </u>

1 . ∠ABC and ∠CBD are a linear pair         Given

2. m∠CBD = m∠EFG                                   Given

3 . ∠ABC and ∠CBD are supplementary   Definition of linear pair

4. m∠ABC + m∠CBD = 180°                        Definition of supplementary

5. m∠ABC + m∠EFG = 180°                         Substitution property

6. ∠ABC and ∠CBD are supplementary    Definition of supplementary

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If 3 oranges and 4 apples cost $14 and 5 oranges and 2 apple cost $21. find the value of 2 oranges and 3 apples.
Semmy [17]

Answer:

$18

Step-by-step explanation:

3 0
3 years ago
Please help. Urgent.
Jet001 [13]

3. ΔPQR ≅ ΔSRT

3. ASA  (Angle - Side - Angle) - we have two triangles where we know two angles and the included side are equal

If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

4. PR ≅ SR

4. ΔPQR ≅ ΔSRT - the corresponding sides are congruent.

7 0
4 years ago
Length of a rectangular prism volume is 2,830.5, width 18.5, height 9
jek_recluse [69]
Well, you get the volume by multiplying the height, the width, and the length. So just work on it backwards.
2,830.5 divided by (9 times 18.5).
9 times 18.5 is 166.5. So that leaves us with 2,830.5 divided by 166.5, which is 17.
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4 0
3 years ago
Read 2 more answers
Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they
Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276

Number of ways of choosing 3 students From all males

n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220

The probability that all are male of choosing '3' students

P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }

P(E) =  \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}

P(E) = 0.067 = 6.71%

<u><em>Final answer</em></u>:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

3 0
4 years ago
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3 years ago
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