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3 years ago

For each function, state weather it’s linear

Mathematics

1 answer:

user100 [1]3 years ago

5 0

Answer:

Step-by-step explanation:

function 1

slope=(1-0)/(0+2)=1/2

or slope=(4-1)/(2-0)=3/2

which is not same.

so it is not linear.

Function 2.

slope=(1+2)/(-3+4)=3/1=3

or slope=(4-1)(-2+3)=3/1=3

or slope=(7-4)/(-1+2)=3/1=3

slope is same. so it is linear.

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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

3 years ago

Consider the expression 5/6(1/2x+6)-3x

Gwar [14]

Ok...I considered it....oh..u want me to answer it...lol..ok

5/6(1/2x + 6) - 3x
5/12x + 30/6 - 3x
5/12x - 36/12x + 5
- 31/12x + 5 <===

3 0

4 years ago

Find the range of the relation below, then determine whether the relation is a function.

Morgarella [4.7K]

Answer:

Step-by-step explanation:

It is a function because it passes the vertical line test and Xs do not repeat, then count out the Y-values which are the range.

4 0

3 years ago

I AM VERY DESPERATE NOW!!! I HAVE ASKED THIS SAME QUESTION 3 TIMES AND NOBODY HAS BEEN ABLE TO ANSWER IT. PLEASE HELP, SOMEONE!!

Alexandra [31]

Answer:

Interest rate is 5.25% a year.

Step-by-step explanation:

r = (1/54)((22252.5/18000) - 1) = 0.004375

r = 0.004375

Convert decimal to percentage

R = 0.004375 * 100 = 0.4375%/month

Calculate annual rate

0.4375%/month × 12 months/year = 5.25% per year.

8 0

3 years ago

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15 Points!!!!! I need help on this :D

nlexa [21]

840 is the answer
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5 0

3 years ago

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