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11 months ago

The graph below shows the same shape was pushing the blue graph

Mathematics

1 answer:

Liula [17]11 months ago

3 0

These are some Rules for the transformations of a function f(x):

a. If the function is shifted left "h" units, then:

$f(x+h)$

b. If the function is shifted right "h" units:

$f(x-h)$

c. If the function is shifted up "h" units:

$f(x)+h$

d. If it is shifted down "h" units:

$f(x)-h$

Notice that the red graph is this function:

$F(x)=x^2$

And both have the same shape, but the blue one is moved to the left 2 units. Therefore the equation of the blue graph is:

$G(x)=(x+2)^2$

THE ANSWER IS OPTION A.

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2 years ago

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

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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)

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Answer:

The function f(x) = 9,000(0.95)^x represents the situation.
After 2 years, the farmer can estimate that there will be about 8,120 bees remaining.
The range values, in the context of the situation, are limited to whole number

Step-by-step explanation:

The "growth" rate is -5%, so the growth factor, the base in the exponential equation, is 1.00-5% =0.95.

Using x=2, we find the population in 2 years is expected to be about ...

f(2) = 9000·0.95^2 ≈ 8123 . . . . about 8120

Using x=4, we find the population in 4 years is expected to be about ...

f(4) = 9000·0.95^4 ≈ 7331 . . . . about 7330

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