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masya89 [10]
3 years ago
6

Ramon is interested in whether the global rise in temperature is also showing up locally in his town, Centerdale. He plans to lo

ok up the average annual temperature for Centerdale for five recent randomly selected years. He wants to report the number of years whose temperature was higher than the previous year’s temperature. What is the random variable in Ramon’s study, and what are its possible values
Mathematics
1 answer:
Nookie1986 [14]3 years ago
6 0

Answer:

explained

Step-by-step explanation:

Ramon wants to report the number of years whose temperature was higher than the previous year’s temperature. This can be done by him as

The random variable is the number of years in which the temperature increased from the previous year.

Its possible values are {0,1,2,3,4,5}.

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Answer:

Arc length MK = 15.45 units (nearest hundredth)

Arc measure = 58.24°

Step-by-step explanation:

Calculate the measure of the angle KLN (as this equals m∠KLM which is the measure of arc MK)

ΔKNL is a right triangle, so we can use the cos trig ratio to find ∠KLM:

\sf \cos(\theta)=\dfrac{A}{H}

where:

  • \theta is the angle
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  • H is the hypotenuse (the side opposite the right angle)

Given:

  • \theta = ∠KLM
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\implies \sf \cos(KLM)=\dfrac{8}{15.2}

\implies \sf \angle KLM=\cos^{-1}\left(\dfrac{8}{15.2}\right)

\implies \sf \angle KLM=58.24313614^{\circ}

Therefore, the measure of arc MK = 58.24° (nearest hundredth)

\textsf{Arc length}=2 \pi r\left(\dfrac{\theta}{360^{\circ}}\right) \quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle)}

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  • r = 15.2
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\implies \textsf{Arc length MK}=2 \pi (15.2)\left(\dfrac{\sf \angle KLM}{360^{\circ}}\right)

\implies \textsf{Arc length MK}=\sf 15.45132428\:units

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