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1 year ago

The radius of a circle is 10 miles. What is the length of a 45° arc?

Mathematics

1 answer:

pshichka [43]1 year ago

6 0

Solution

We are given

Radius (r) = 10 miles

Angle (theta) = 45 degrees

We want to find the length of the arc

Note: Formula for the length of an Arc

$\begin{gathered} l=\frac{\theta}{360}\times2\pi r \\ length\text{ }Of\text{ }Arc=\frac{45}{360}\times2\times\pi\times10 \\ length\text{ }Of\text{ }Arc=\frac{1}{8}\times20\pi \\ length\text{ }Of\text{ }Arc=\frac{5}{2}\pi \\ length\text{ }Of\text{ }Arc=7.853981634 \\ length\text{ }Of\text{ }Arc=7.854miles \end{gathered}$

Therefore, the length of the arc is

$\begin{equation*} 7.854miles \end{equation*}$

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0.6127 = 61.27% probability of selling no more than 6 properties in one week.

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For each property, there are only two possible outcomes. Either it is sold, or it is not. The probability of a property being sold is independent of any other property. This means that the binomial probability distribution is used to solve this question.

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The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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This means that $n = 12$

She feels that there is a 50% chance of selling any one property during a week.

This means that $p = 0.5$

Compute the probability of selling no more than 6 properties in one week.

This is:

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

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$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.2256$

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0.6127 = 61.27% probability of selling no more than 6 properties in one week.

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