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monitta
3 years ago
9

Please help with either of these questions. Put the answer in the comments please

Mathematics
1 answer:
Lena [83]3 years ago
6 0

Answer:

2) distance = 4 km

Displacement = 4 km east

3) distance = 35 km

Displacement = 5 km north

Step-by-step explanation:

2. He runs for 3 km in the east direction and then stops.. He now runs another 1 km in the same direction then stops.. Thus,

Total distance covered = 3 + 1 = 4 km

Total displacement is how far he is from his starting point = 3 + 1 = 4 km east

3) Runs 20km north and then runs 15 km back from starting point.

Thus;

Total distance covered = 20 + 15 = 35 km

Total displacement is distance from starting point = 20 - 15 = 5 km north

Subtracted because an opposite direction was taken after the 20km back to her starting point.

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Zepler [3.9K]
\frac{-7}{12} = \frac{y}{-36}

First, simplify \frac{y}{-36} to - \frac{y}{36} / Your problem should look like: - \frac{7}{12} = -\frac{y}{36}
Second, multiply both sides by 36. / Your problem should look like: -21=-y
Third, multiply both sides by -1. / Your problem should look like: 21 = y
Fourth, switch your sides. / Your problem should look like: y = 21

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5 0
3 years ago
Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%. Find
Papessa [141]

Answer:

1) \text{P(at least one boy and one girl)}=\frac{3}{4}

2) \text{P(at least one boy and one girl)}=\frac{3}{8}

3) \text{P(at least two girls)}=\frac{1}{2}

Step-by-step explanation:

Given : Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%.

To  Find : The probability of each event.  

1) P(at least one boy and one girl)

2) P(two boys and one girl)

3) P(at least two girls)        

Solution :

Let's represent a boy with B and a girl with G

Mr. and Mrs. Romero are expecting triplets.

The possibility of having triplet is

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Total outcome = 8

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

1) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, BGG, GBB, GBG, GGB=6

\text{P(at least one boy and one girl)}=\frac{6}{8}

\text{P(at least one boy and one girl)}=\frac{3}{4}

2) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, GBB=3

\text{P(at least one boy and one girl)}=\frac{3}{8}

3) P(at least two girls)

Favorable outcome = BGG, GBG, GGB, GGG=4

\text{P(at least two girls)}=\frac{4}{8}

\text{P(at least two girls)}=\frac{1}{2}

4 0
3 years ago
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Answer:

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