All categories

2 years ago

1. Mr.Rodriguez Works at a used car dealership and earns a base pay of $1,000 per

Mathematics

1 answer:

Olin [163]2 years ago

4 0

Answer:

20%

Step-by-step explanation:

all you do is think of the first 2 numbers and say that is the percent.

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

2 years ago

Evaluate the expression 2bc^2 for b = 1.5 and c = 5.

salantis [7]

Answer:

108

Step-by-step explanation:

2(1.5)(6^2)=3(36)=108

3 0

2 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

storchak [24]

Answer:

$a_n = 1200(1.35)^n$

Step-by-step explanation:

Equation to remember:

$P_0(1+r)^t$ where t = time

$P_0$ = 1200 since that's the initial population

$r$ = .35 rate of change \\

$t$ = 5 years

Plug in the numbers;

$a_5 = 1200(1+.35)^5$

Since the prompt looks like it's asking for an equation given any number of years.. then just change 5 to n for number of years.

$a_n = 1200(1.35)^n$

8 0

2 years ago

Read 2 more answers

Guillermo saved 25% of his paycheck last week. His paycheck was $1000. How much money did he save?

Bingel [31]

He saved 250 dollars.

8 0

2 years ago

Read 2 more answers

The general solution of 2 y ln(x)y' = (y^2 + 4)/x is

Sav [38]

Replace $y'$ with $\dfrac{\mathrm dy}{\mathrm dx}$ to see that this ODE is separable:

$2y\ln x\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2+4}x\implies\dfrac{2y}{y^2+4}\,\mathrm dy=\dfrac{\mathrm dx}{x\ln x}$

Integrate both sides; on the left, set $u=y^2+4$ so that $\mathrm du=2y\,\mathrm dy$ ; on the right, set $v=\ln x$ so that $\mathrm dv=\dfrac{\mathrm dx}x$ . Then