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hodyreva [135]
2 years ago
10

The base of a ladder is placed 2.8m away from a tree that is 4.5m tall. The end of the ladder touches the top of the tree. Calcu

late the length of the ladder.
Mathematics
1 answer:
romanna [79]2 years ago
5 0

The length of the ladder is 5.3 meter if the base of a ladder is placed 2.8 m away from a tree that is 4.5 m tall. The end of the ladder touches the top of the tree.

<h3>What is a right-angle triangle?</h3>

It is a triangle in which one of the angles is 90 degrees and the other two are sharp angles. The sides of a right-angled triangle are known as the hypotenuse, perpendicular, and base.

As per the picture attached, the triangle is a right angle triangle:

Let's assume the length of the ladder is x meter

From the Pythagoras theorem:

x = √(2.8²+4.5²)

x = √(7.84+20.25)

x = 5.3 m

Thus, the length of the ladder is 5.3 meter if the base of a ladder is placed 2.8 m away from a tree that is 4.5 m tall. The end of the ladder touches the top of the tree.

Learn more about the right angle triangle here:

brainly.com/question/3770177

#SPJ1

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Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

A_{i}: the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

P(A_{i})=0.1 for i = 1, 2, 3, 4, 5

The complement for A_{i} is given by

A_{i}^{$c$}: the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P(A_{i}^{$c$})=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})=(0.9)^{5}=0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P(A_{1}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}∩A_{4}^{$c$}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}∩A_{5}^{$c$})+P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5})=5×(0.1)×(0.9)^{4}=0.32805

because we have independent events.

(c) The probability that at least one contains high levels of contamination is

P(A_{1}∪A_{2}∪A_{3}∪A_{4}∪A_{5})=1-P(A_{1}^{$c$}∩A_{2}^{$c$}∩A_{3}^{$c$}∩A_{4}^{$c$}∩A_{5}^{$c$})=1-0.59049=0.40951

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Step-by-step explanation:

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