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ElenaW [278]
2 years ago
15

The sum of two positive integers, a and b, is at least 30. The difference of the two integers is at least 10. If b is the greate

r integer, which system of inequalities could represent the values of a and b?
a + b ≥ 30 b ≥ a + 10
a + b ≥ 30 b ≤ a – 10
a + b ≤ 30 b ≥ a + 10
a + b ≤ 30 b ≤ a – 10
Mathematics
2 answers:
KIM [24]2 years ago
8 0

a + b ≥ 30,  b ≥ a + 10, the system of inequalities could represent the values of a and b

option A

<u>Step-by-step explanation:</u>

Here we have , The sum of two positive integers, a and b, is at least 30. The difference of the two integers is at least 10. If b is the greater integer, We need to find which system of inequalities could represent the values of a and b . Let's find out:

Let two numbers be a and b where b>a . Now ,

  • The sum of two positive integers, a and b, is at least 30

According to the given statement we have following inequality :

⇒ a+b\geq 30

  • The difference of the two integers is at least 10

According to the given statement we have following inequality :

⇒ b-a\geq 10

⇒ b-a+a\geq 10 +a

⇒ b\geq 10 +a

Therefore , Correct option is A) a + b ≥ 30,  b ≥ a + 10

prohojiy [21]2 years ago
7 0

Answer:

Start with the difference between the two integers

If b is greater than a, then we can write 'b - a ≥ 10' or 'b ≥ a + 10'

Then we have 'b + a ≥ 30'

Answer: b + a ≥ 30 and b ≥ a + 10

Option A

Step-by-step explanation:

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Answer:

Question 1: P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16}}{ \frac{4}{16}} = \frac{1}{2}

Question 2:

A. P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}

B. P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}

C. P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}

Step-by-step explanation:

Conditional probability is defined by

P(A|B)= \frac{P(A and B)}{P(B)}

with P(A and B) beeing the probability of both events occurring simultaneously.

Question 1:

B: Baseball League Championships won, beeing

P ( B ) = \frac{ 6 }{16}

Y: Championships won by the 10 - 12 years old, beeing

P ( Y)= \frac{ 4 }{ 16 }

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16} }{ \frac{4}{16} }  = \frac{1}{2}

Question 2.A:

Y: Championships won by the 10 - 12 years old, beeing

P ( Y)= \frac{ 4 }{ 16 }

B: Baseball League Championships won, beeing

P ( B ) = \frac{ 6 }{16}

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}

Question 2.B:

Z: Championships won by the 13 - 15 years old, beeing

P ( Z)= \frac{ 1 }{ 16 }

B: Baseball League Championships won, beeing

P ( B ) = \frac{ 6 }{16}

then

P( Z and B)= \frac{ 1 }{ 16 }[/tex]

By definition,

P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}

Question 3.B

Y: Championships won by the 10 - 12 years old, beeing

P ( Y)= \frac{ 4 }{ 16 }

Z: Championships won by the 13 - 15 years old, beeing

P ( Z)= \frac{ 1 }{ 16 }

then

P (Y or Z) = P(Y) + P(Z) = \frac{6}{16}

B: Baseball League Championships won, beeing

P ( B ) = \frac{ 6 }{16}

so

P((YorZ) and B)= \frac{3}{16}

By definition,

P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}

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