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aleksandr82 [10.1K]

3 years ago

6

Every month, $60 is withdrawn from Tom's savings account to pay for his gym membership. He has enough savings to withdraw no mor

e than $240. For how many months can Tom pay for his gym membership?

1 answer:

IrinaVladis [17]3 years ago

7 0

Answer: 9 month

Step-by-step explanation:

1. 35x < 315 = let x be the number of month

2. x < 9 = divide both sides by 35

result. 9 month

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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

Which represents the solution set of 5(x+5)<85

andrey2020 [161]

X<12
<span>5(x+5)<85 </span>
<span>5(x + 5)/5 < 85/5 </span>
<span>x + 5 < 17 </span>
<span>x + 5 - 5 < 17 - 5</span>

3 0

3 years ago

About 5% of the population has a particular genetic mutation. 300 people are randomly selected.

xz_007 [3.2K]

The standard deviation for the number of people with the genetic mutation is 3.77

<h3>How to determine the standard deviation?</h3>

The given parameters are:

Sample size, n = 300

Proportion that has the particular genetic mutation, p = 5%

The standard deviation for the number of people with the genetic mutation is calculated as:

Standard deviation = √np(1 - p)

Substitute the known values in the above equation

Standard deviation = √300 * 5% * (1 - 5%)

Evaluate the product

Standard deviation = √14.25

Evaluate the exponent

Standard deviation = 3.77

Hence, the standard deviation for the number of people with the genetic mutation is 3.77

Read more about standard deviation at

brainly.com/question/12402189

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7 0

2 years ago

81(p+q)^2-9p-9q <br> Factor this

polet [3.4K]

Answer:

9(p + q)(9p + 9q - 1)

Step-by-step explanation:

Given

81(p + q)² - 9p - 9q ← factor out - 9 from these 2 terms

= 81(p + q)² - 9(p + q) ← factor out 9(p + q) from each term

= 9(p + q)(9(p + q) - 1)

= 9(p + q)(9p + 9q - 1)

8 0

3 years ago

Read 2 more answers

The cube root of the difference of a number and 2

julia-pushkina [17]

Answer:

The difference of two cubes is equal to the difference of their cube roots times a trinomial, which contains the squares of the cube roots and the opposite of the product of the cube roots. A number's opposite is that same number with a different sign in front.

Step-by-step explanation:

8 0

3 years ago