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1 year ago

Statement "if a and b are both positive integer numbers, then a +b > a

Mathematics

1 answer:

bezimeni [28]1 year ago

3 0

Answer:

Always

Step-by-step explanation:

Subtracting a from both sides yields b>0, which is always true since b is a positive integer.

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Translate Each sentence into an equation!!!

Amanda [17]

20+2N=52
now subtract 20 from each side, to get 2N alone
2N=32
now divide each side by 2
N=16

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3 years ago

What is the quotient of 2 1/5 ÷ 6 3/5 A. 1/3 B. 3/4 C. 24/35 D. 2/3

zepelin [54]

Answer:

A.1/3

Step-by-step explanation:

on the picture

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2 years ago

I need some help!!!!!!!!!!!!!!!!!!!!!!!

vampirchik [111]

Step-by-step explanation:

We are dealing with theoretical probability. This means that our probability of said outcome is

$\frac{x}{y}$

where x is the outcome we want, and y is the total number of outcomes possible.

28. There is 3 triangles out of 6 polygons so the probability is 1/2.

29.There is 1 Pentagon so the probability is 1/6.

30. This is a complement to question 28. A complement is the inverse of the problem. A complement and its original add to 1 so the probability of both getting a triangle is 1/2.

31. There is 4 non quadraletrial so the probability is 2/3.

32. There is 3 figures that has more sides than three so 1/2 is the probability.

33.There is only multi right angles figures so 1/2 is the probability.

34. There is 30 days in April so the probability it's the 29th is

$\frac{1}{30}$

35. There is 31 days in July and 15 days after the 16th. So the probability

$\frac{15}{31}$

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2 years ago

A small pebble has a mass of about 20 L b. 20 ml 20 g 20 kg d.

zhuklara [117]

Answer:

20g

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7 0

3 years ago

PLEASE HELP! The table shows the number of championships won by the baseball and softball leagues of three youth baseball divisi

Irina18 [472]

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

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

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

By definition,

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

3 0

3 years ago