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AnnyKZ [126]
2 years ago
15

For a project in his Geometry class, Mamadou uses a mirror on the ground to measure the height of his school's football goalpost

. He walks a distance of 13.75 meters from his school, then places a mirror on flat on the ground, marked with an X at the center. He then steps 2.6 meters to the other side of the mirror, until he can see the top of the goalpost clearly marked in the X. His partner measures the distance from his eyes to the ground to be 1.75 meters . How tall is the goalpost? Round your answer to the nearest hundredth of a meter .

Mathematics
1 answer:
andreev551 [17]2 years ago
4 0

Answer:

Height of the goalpost is 9.25 m.

Step-by-step explanation:

As per the rule of reflection in physics,

Angle of incidence = Angle of reflection

As we can see in the picture attached, both the angles (θ) are equal.

m∠ACB = m∠ECD = 90° - θ

m∠ABC = m∠ADC = 90°

Therefore, both the triangles ΔABC and ΔEDC will be similar.

And by the property of similar triangles, their corresponding sides will be proportional.

\frac{AB}{ED}= \frac{CB}{CD}

\frac{1.75}{ED}=\frac{2.6}{13.75}

ED = \frac{1.75\times 13.75}{2.6}

ED = 9.25 m

Height of the goalpost is 9.25 m.

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Answer:

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Step-by-step explanation:

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Solution of the system of equations will be the point of intersection of these lines.

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3 years ago
Louisa has 4 yards of fabric. She uses 2/5 yard of fabric to make a doll dress. How many doll dresses can Louisa make with the f
Katyanochek1 [597]

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I believe the answer is D. Number line from 0 to 4 is divided into 20 equal parts. Number line is grouped into 10 equal parts.

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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)

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