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Nonamiya [84]
3 years ago
11

A plane intersects the prism perpendicular to the base, intersecting opposite sides of the base. Which best describes the cross

section?
Mathematics
1 answer:
hammer [34]3 years ago
3 0

Answer:

Step-by-step explanation:

For a rectangular prism, a plane intersecting this prism produces a two dimensional figure as its cross section and in this case the cross section is a rectangle.

For a triangular prism, a plane intersecting this prism produces a two dimensional figure as its cross section and in this case the cross section is a triangle.

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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
[Quick Answer Needed] Which of the following shows the extraneous solution to the logarithmic equation?
Nitella [24]

Answer:

C

Step-by-step explanation:

Given the logarithmic equation

\log_4x+\log_4(x-3)=\log_4(-7x+21)

First, notice that

x>0\\ \\x-3>0\Rightarrow x>3\\ \\-7x+21>0\Rightarrow 7x

So, there is no possible solutions, all possible solutions will be extraneous.

Solve the equation:

\log_4x+\log_4(x-3)=\log_4x(x-3),

then

\log_4x(x-3)=\log_4(-7x+21)\\ \\x(x-3)=-7x+21\\ \\x^2-3x+7x-21=0\\ \\x^2+4x-21=0\\ \\D=4^2-4\cdot 1\cdot (-21)=16+84=100\\ \\x_{1,2}=\dfrac{-4\pm 10}{2}=-7,\ 3

Hence, x=3 and x=-7 are extraneous solutions

6 0
3 years ago
400g of rapberries and 300g of strawberries cost £7.46. 500g of strawberries cost £4.10 work out the cost of 300g of raspberries
MAXImum [283]

Answer: £5.39

Step-by-step explanation:

500g strawberries = 4.10

100 g = 0.82

300g = 2.46

400g raspberries = 7.46 - 2.46 = 5.00

100g = 1.25

300g = 3.75

200g strawberries = 0.82 * 2 = 1.64

3.75 + 1.64 = 5.39

8 0
3 years ago
Integer problems that equal 36
klio [65]
What integer is less than 37 but greater than 35?
5 0
3 years ago
Multiple choice please answer please
eduard

Answer: 33.85 mm

Step-by-step explanation:

Use trigonometry, tan equation

You are give the tan angle and adjacent side, you need to find the opposite side so

O

T A

Tan(72)= x/11

x= tan(72) x 11

x= 33.85

8 0
3 years ago
Read 2 more answers
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