All categories

3 years ago

The table below shows the commission for selling cameras in a

Mathematics

1 answer:

barxatty [35]3 years ago

5 0

Answer:

Step-by-step explanation:

You first do 1,110-150= 960 dollars.

Then you divide 960 by 15 dollars = 64 cameras

So the first 20 cameras plus the 64 cameras equal 84 cameras.

You might be interested in

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

3. (5 + 2) = (3 . 5) + (3 . 2) *

UNO [17]

Answer:

C) Distributive Property

Step-by-step explanation:

You're <em>distributing</em> in the 3 when you multiply it by the 5 and the 2 in the algebraic process:

➦ ➦

3•(5 + 2) = (3 • 5) + (3 • 2)

So, the 3 was <em>distributed</em> between the 5 and 2 when you multiplied.

Hence, this is the distributive property.

8 0

3 years ago

What is 82×68 this is very hard

zubka84 [21]

Answer:

5576

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Solve for the variable: 3x + 3 = 36 11 12 13 33

Andrei [34K]

The Answer Is A)11

Explanation: Subtract 3 From Both Sides By 3 Then Divide By 3

3 0

3 years ago

Read 2 more answers

How many cm more than 2 m is the perimeter of a regular hexagon of side 35 cm?

bearhunter [10]

A Hexagon Is A 6 Sided Figure. So First, We Need To Know How Many Centimeters Are In A Meter. There Are 100 Centimeters In A Meter. Next, Do 2 Meters * 100 centimeters To Get 2 Meters Is Equal To 200 Centimeters. Next, We Do 35*6. We Get 210. Now, Do The Subtraction. 210-200 = 10cm. The Regular Hexagon's Perimeter Is 10cm Larger. I Hope I Helped! :D

3 0

3 years ago