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3 years ago

9 is the difference between two numbers, and their sum is 31. Find the numbers.

Mathematics

1 answer:

algol [13]3 years ago

4 0

Answer:

The first number is 20, the second one is 11

Step-by-step explanation:

Let the first number be x and the second number be y

Since it says "9 is the difference between two numbers", the first equation will be x-y=9

It also says that "their sum is 31". That means the second equation will be x+y=31

here are the two equations:

x-y=9

x+y=31

You can solve this any way you like, but I'll show elimination

add the two equations

x-y=9

x+y=31

2x=40

divide by 2

x=20

Substitute 20 as x in one of the equations

20+y=31

subtract 20 from both sides

y=11

That means the two numbers are 20 and 11

Hope this helps!

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3 years ago

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

Step-by-step explanation:

Given:

Total number of students in the class = 30

Number of students who plays basket ball = 19

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)

where

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

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

$P(A \cup B)= \frac{31}{30} -\frac{8}{30}$

$P(A \cup B)= \frac{23}{30}$

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