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3 years ago

7

What natural disaster would pose the most serious threat to buildings were

1 answer:

ryzh [129]3 years ago

5 0

Answer:

The answer is "floods"

Step-by-step explanation:

It is a natural phenomenon or occurrence if an area of land, which is typically desert country immediately gets dissolved covered in water.
There are several other floods, that suddenly and a wide gap, the other will take weeks or even months to construct and release.
This cause is the most major threat to building structures, in which New York City to first be destroyed.

You might be interested in

Samin can run 5 kilometres in 30 minutes. Assuming she keeps a constant pace, how many kilometres can she run in 45 minutes?

Vladimir [108]

Answer:

7.5

Step-by-step explanation: She can run 5km in 30 mins. That means she can run 1 km every 6 mins. 45/6= 7.5. She can run 7.5 km in 45 mins. Such an amazing feat my chubbiness can't compete with.

7 0

2 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

2 years ago

Jerome is painting a rectangular toolbox that is 20 inches by 10 inches by 8 inches. A tube of paint covers 300 square inches. W

Valentin [98]

Answer:

880 in³

Step-by-step explanation:

L = 20, W = 10, H = 8

2(h × W) + 2(h × L) + 2(W × L)

= 2(8*10) + 2(8*20) + 2(10*20)

= 2(80) + 2(160) + 2(200)

= 160 + 320 + 400

= 880 in³

8 0

3 years ago

HELP !!!! picture is shown GEOMETRY

kenny6666 [7]

The right triangle is also Isosceles. It has two equal angles of 45°.

(hypotenuse)² = 2S²

4 = 2S² => S = √2

Therefore B is the correct answer.

8 0

2 years ago

X^2+9x+20=0<br><br> What are the steps to solve by factoring?

Dimas [21]

First simplify the equation and next multiply the 20 with the x^2 and find a number that adds together to give you the middle

6 0

3 years ago

Read 2 more answers