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konstantin123 [22]

3 years ago

13

Question 4 Multiple Choice Worth 4 points)

1 answer:

belka [17]3 years ago

3 0

Answer:

y = 1/2x+11

Step-by-step explanation:

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What is the surface area of the right rectangular prism?

garri49 [273]

Answer:

Hello! answer:

Step-by-step explanation:

6 × 12 = 72

6 × 12 = 72

3 × 12 = 36

3 × 12 = 36

3 × 6 = 18

3 × 6 = 18

72 + 72 + 36 + 36 + 18 + 18 = 252 therefore the surface area would be 252 hope that helps!

6 0

2 years ago

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Is 3x-5+y=2y-4 linear if so write the equation in standard form and give the values of A B and C

ruslelena [56]

Equation in standard form:
Ax+By=C

We have:
3x-5+y=2y-4
3x+y-2y=-4+5
3x-y=1

Therefore:
A=3
B=-1
C=1

Answer: the equation in standard form would be: 3x-y=1; and the values of A, B and C would be: A=3; B=-1 and C=1.

8 0

3 years ago

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

A cube has a length, width and height of 6 feet. What is the volume in cubic INCHES?

Inessa05 [86]

Answer:

Step-by-step explanation:

4 0

3 years ago

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What is the input value other than 7 for which g(x)=4? WILL GIVE BRAINLIEST ANSWER!

trasher [3.6K]

-8 the point on the line with Y=4

6 0

3 years ago