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3 years ago

Let ƒ(x) = x^2 + 2x and g(x) = 2x. Evaluate the composition (ƒ ∘ g)(2).

Mathematics

1 answer:

FinnZ [79.3K]3 years ago

7 0

Answer:

The answer is option D.

Step-by-step explanation:

f(x) = x² + 2x

g(x) = 2x

To find (ƒ ∘ g)(2) first find (ƒ ∘ g)

To find (ƒ ∘ g) wherever you see x in f(x) replace it with g(x)

That's

(ƒ ∘ g) = (2x)² + 2(2x)

= 4x² + 4x

Substitute 2 into the expression

That's

(ƒ ∘ g)(2) = 4(2)² + 4(2)

= 4(4) + 8

= 16 + 8

= 24

Hope this helps you.

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For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.

nordsb [41]

Answer:

<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>

By De morgan's law

$(A\cap B)^{c}=A^{c}\cup B^{c}\\\\P((A\cap B)^{c})=P(A^{c}\cup B^{c})\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq 1-P(A)+1-P(B)\\\\-P(A\cap B)\leq 1-P(A)-P(B)\\\\P(A\cap B)\geq P(A)+P(B)-1$

which is Bonferroni’s inequality

<h3>Result 1: P (Ac) = 1 − P(A)</h3>

Proof

If S is universal set then

$A\cup A^{c}=S\\\\P(A\cup A^{c})=P(S)\\\\P(A)+P(A^{c})=1\\\\P(A^{c})=1-P(A)$

<h3>Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B) and P(A) ≥ P(B)</h3>

Proof:

If S is a universal set then:

$A\cup(B\cap A^{c})=(A\cup B) \cap (A\cup A^{c})\\=(A\cup B) \cap S\\A\cup(B\cap A^{c})=(A\cup B)$

Which show A∪B can be expressed as union of two disjoint sets.

If A and (B∩Ac) are two disjoint sets then

$P(A\cup B) =P(A) + P(B\cap A^{c})---(1)\\$

B can be expressed as:

$B=B\cap(A\cup A^{c})\\$

If B is intersection of two disjoint sets then

$P(B)=P(B\cap(A)+P(B\cup A^{c})\\P(B\cup A^{c}=P(B)-P(B\cap A)$

Then (1) becomes

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

<h3>Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)</h3>

Proof:

If A and B are two disjoint sets then

$A=A\cap(B\cup B^{c})\\A=(A\cap B) \cup (A\cap B^{c})\\P(A)=P(A\cap B) + P(A\cap B^{c})\\$

<h3>Result 4: If B ⊂ A, then A∩B = B. Therefore P (A)−P (B) = P (A ∩ Bc) </h3>

Proof:

If B is subset of A then all elements of B lie in A so A ∩ B =B

$A =(A \cap B)\cup (A\cap B^{c}) = B \cup ( A\cap B^{c})$

where A and A ∩ Bc are disjoint.

$P(A)=P(B\cup ( A\cap B^{c}))\\\\P(A)=P(B)+P( A\cap B^{c})$

From axiom P(E)≥0

$P( A\cap B^{c})\geq 0\\\\P(A)-P(B)=P( A\cap B^{c})\\P(A)=P(B)+P(A\cap B^{c})\geq P(B)$

Therefore,

P(A)≥P(B)

8 0

2 years ago

Find the value(s) of "k" that will cause the equation 4x^2 +kx + 4 to have one real solution.

Andreas93 [3]

Answer: k = {8, -8}

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

In order for a quadratic equation to have exactly one solution, the discriminant must equal zero. → b² - 4ac = 0

4x² + kx + 4 = 0

↓ ↓ ↓

a=4 b=k c=4

b² - 4ac = 0

k² - 4(4)(4) = 0

k² = 64

k = √64

k = ± 8

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3 years ago

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Veronika [31]

Mp is congruent to rs

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wolverine [178]

Answer:

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2 years ago

A baseball team won 2 games. They played 17 games total. Which ratio shows how many games they lost

Contact [7]

Answer:

15/17

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if they won 2 games they lost 15 (17-2) so the ratio is 15/17

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2 years ago