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3 years ago

5/6 +3/7 give answer as a mixed number

Mathematics

2 answers:

Lyrx [107]3 years ago

5 0

Answer:

$1\frac{11}{42}$

Step-by-step explanation:

$\frac{5}{6} = \frac{35}{42}$
$\frac{3}{7} = \frac{18}{42}$
$\frac{35}{42} + \frac{18}{42} = \frac{53}{42}$
$\frac{53}{42} = 1\frac{11}{42}$

I hope this helps!

Bogdan [553]3 years ago

5 0

Answer:

56/42, or 1 14/42

Step-by-step explanation:

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In a certain Algebra 2 class of 30 students, 19 of them play basketball and 12 of them play baseball. There are 8 students who p

Alenkinab [10]

Answer:

Probability that a student chosen randomly from the class plays basketball or baseball is $\frac{23}{30}$ or 0.76

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Given:

Total number of students in the class = 30

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Number of students who plays base ball = 12

Number of students who plays base both the games = 8

To find:

Probability that a student chosen randomly from the class plays basketball or baseball=?

Solution:

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$ ---------------(1)

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P(A) = Probability of choosing a student playing basket ball

P(B) = Probability of choosing a student playing base ball

P(A \cap B) = Probability of choosing a student playing both the games

Finding P(A)

P(A) = $\frac{\text { Number of students playing basket ball }}{\text{Total number of students}}$

P(A) = $\frac{19}{30}$ --------------------------(2)

Finding P(B)

P(B) = $\frac{\text { Number of students playing baseball }}{\text{Total number of students}}$

P(B) = $\frac{12}{30}$ ---------------------------(3)

Finding $P(A \cap B)$

P(A) = $\frac{\text { Number of students playing both games }}{\text{Total number of students}}$

P(A) = $\frac{8}{30}$ -----------------------------(4)

Now substituting (2), (3) , (4) in (1), we get

$P(A \cup B)= \frac{19}{30} + \frac{12}{30} -\frac{8}{30}$

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