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3 years ago

Help is all I need..

Mathematics

2 answers:

Solnce55 [7]3 years ago

7 0

Answer:

${(x - 1)}^{2} = 16 \\ x - 1 = |4| \\ x = |4| + 1$

x = +4 +1

x = +5

and

x = -4 + 1

x = -3

Sladkaya [172]3 years ago

5 0

The answer is 5, enjoy

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The domain of discourse is sets. Suppose some predicate Q is true for all elemenst of A: ∀a ∈ A Q(a). Recall the definition of s

Sloan [31]

Answer:

See the step-by-step explanation

Step-by-step explanation:

Let c be any element of C. (I'm not sure wether you have to assume that C is non-empt or not)

C is a subset of B. That means that as c is in C, it is also in B. ( $c \in C \Rightarrow c \in B$ )

Now, B is a subset of A. It follows that as $c \in B \Rightarrow c \in A$ .

That means c is an element of A. The predicate Q is true for all elements of A, including c.

Because we let c be any element of C, we have proven that the predicate Q is true for all elements in C.

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3 years ago

The box plots below show attendance at a local movie theater and high school basketball games: Two box plots are shown. The top

nignag [31]

Answer:

Inter-quartile range

Step-by-step explanation:

When you have the quartile, you can use the IQR (interquartile range) to measure the dispersion.

IQR = Q3 - Q1

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3 years ago

1/5(x+2)+2x=6x-10 what does x=

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3 years ago

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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