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3 years ago

How do you solve 15x1.12

Mathematics

2 answers:

Andru [333]3 years ago

6 0

Put 1.12 on top and times regullarly you should get 1280.00

solniwko [45]3 years ago

5 0

16.8 but you can also use an app called photo math that solves it out for 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

3 years ago

Arlene buys a phone case and a charging cord for 15% off. The original price of the phone case is 18$. Her total discount is 4.2

nordsb [41]

Answer:13.80

Step-by-step explanation:18-4.20=13.80

6 0

2 years ago

Write as a decimal. Be sure to put the zero before the decimal point. 17.9%

nlexa [21]

0.179

When it wants a percent to a decimal just move the decimal 2 points to the left.

7 0

3 years ago

What is the value of x?<br> edge2020

mezya [45]

Answer:

95⁰

Step-by-step explanation:

next time you want to figure this out, here's how

- they are supplementary angles, so you just subtract 45⁰ from 180⁰

180-45=95

so, you would get the answer 95⁰

for next time, you might not have as much information as you do in this equation so be careful

hope this helps!

7 0

3 years ago

Read 2 more answers

A student factors a^6 - 64 to (a^2-4)(a^4 + 4a^2+ 16).

Neko [114]

Answer:

Option (b) is correct.

The expression is equivalent, but the term is not completely factored.

Step-by-step explanation:

edge 2020

4 0

3 years ago