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3 years ago

In a certain Algebra 2 class of 21 students, 10 of them play basketball and 9 of them

Mathematics

1 answer:

Oliga [24]3 years ago

5 0

Step-by-step explanation:

out of the 21 students, 8 do not play neither basketball nor baseball.

so, 13 students play at least one of these games.

10 play basketball, and 9 play baseball. that would be 19 together. but we have only 13 players.

that means 6 students (19 - 13) must play both games.

remember, a probability is always

desired cases / possible cases

so, when we are looking for the students, who play both games, we have 6 desired cases out of a total of 21.

and the probability to pick one of them is therefore

6/21 = 2/7 = 0.285714286...

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Answer:

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Step-by-step explanation:

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4 years ago

The shadow of a tower at a time is three times as long as its shadow when the angle of elevation of the Sun is 60°. Find the ang

Naddika [18.5K]

Answer:

$30^{\circ}$ .

Step-by-step explanation:

Let $\theta$ denote the unknown angle of elevation. Let $h$ denote the height of the tower.

Refer to the diagram attached. In this diagram, ${\sf A}$ denotes the top of the tower while ${\sf B}$ denote the base of the tower; ${\sf BC}$ and ${\sf BD}$ denote the shadows of the tower when the angle of elevation of the sun is $60^{\circ}$ and $\theta$ , respectively. The length of segment ${\sf AB}$ is $h$ ; $\angle {\sf ACB} = 60^{\circ}$ , $\angle {\sf ADB} = \theta$ , and ${\sf BD} = 3\, {\sf BC}$ ..

Note that in right triangle $\triangle {\sf ABC}$ , segment ${\sf AB}$ (the tower) is opposite to $\angle {\sf ACB}$ . At the same time, segment ${\sf BC}$ (shadow of the tower when the angle of elevation of the sun is $60^{\circ}$ ) is adjacent to $\angle {\sf ACB}$ .

Thus, the ratio between the length of these two segments could be described with the tangent of $m\angle {\sf ACB} = 60^{\circ}$ :

$\begin{aligned}\tan(\angle {\sf ACB}) &= \frac{\text{opposite}}{\text{adjacent}} = \frac{{\sf AB}}{{\sf BC}}\end{aligned}$ .

$\begin{aligned}\frac{{\sf AB}}{{\sf BC}} = \tan(60^{\circ}) = \sqrt{3}\end{aligned}$ .

The length of segment ${\sf AB}$ is $h$ . Rearrange this equation to find the length of segment ${\sf BC}$ :

$\begin{aligned} {\sf BC} &= \frac{{\sf AB}}{\tan(\angle ACB)} \\ &= \frac{h}{\tan(60^{\circ})}\\ &= \frac{h}{\sqrt{3}} \\ &\end{aligned}$ .

Therefore:

$\begin{aligned}{\sf BD} &= 3\, {\sf BC} \\ &= \frac{3\, h}{\sqrt{3}} \\ &= (\sqrt{3})\, h\end{aligned}$ .

Similarly, in right triangle ${\sf ABD}$ , segment ${\sf AB}$ (the tower) is opposite to $\angle {\sf ADB}$ . Segment ${\sf BD}$ (shadow of the tower, with $\theta$ as the angle of elevation of the sun) is adjacent to $\angle {\sf ADB}$ .