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Sonbull [250]
3 years ago
11

Your waiter at a restaurant suggests you leave a tip of $10 on a $50 bill. What percentage is that?

Mathematics
1 answer:
Bezzdna [24]3 years ago
6 0

Answer:

20%

Step-by-step explanation:

To find what percentage is $10 out of $50, we divide 10 by 50:

\frac{10}{50}=0.2

If we want to get the result in percentage form, we simply multiply it by 100%:

0.2\cdot 100\%=20\%

So a tip of $10 on a $50 bill is a tip of 20%.

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Two lighthouses are located 75 miles from one another on a north-south line. If a boat is spotted S 40o E from the northern ligh
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Answer:

The northern lighthouse is approximately 24.4\; \rm mi closer to the boat than the southern lighthouse.

Step-by-step explanation:

Refer to the diagram attached. Denote the northern lighthouse as \rm N, the southern lighthouse as \rm S, and the boat as \rm B. These three points would form a triangle.

It is given that two of the angles of this triangle measure 40^{\circ} (northern lighthouse, \angle {\rm N}) and 21^{\circ} (southern lighthouse \angle {\rm S}), respectively. The three angles of any triangle add up to 180^{\circ}. Therefore, the third angle of this triangle would measure 180^{\circ} - (40^{\circ} + 21^{\circ}) = 119^{\circ} (boat \angle {\rm B}.)

It is also given that the length between the two lighthouses (length of \rm NS) is 75\; \rm mi.

By the law of sine, the length of a side in a given triangle would be proportional to the angle opposite to that side. For example, in the triangle in this question, \angle {\rm B} is opposite to side \rm NS, whereas \angle {\rm S} is opposite to side {\rm NB}. Therefore:

\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of NB}}{\sin(\angle {\rm S})} \end{aligned}.

Substitute in the known measurements:

\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of NB}}{\sin(21^{\circ})} \end{aligned}.

Rearrange and solve for the length of \rm NB:

\begin{aligned} & \text{length of NB} \\ =\; & (75\; \rm mi) \times \frac{\sin(21^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 30.73\; \rm mi\end{aligned}.

(Round to at least one more decimal places than the values in the choices.)

Likewise, with \angle {\rm N} is opposite to side {\rm SB}, the following would also hold:

\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of SB}}{\sin(\angle {\rm N})} \end{aligned}.

\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of SB}}{\sin(40^{\circ})} \end{aligned}.

\begin{aligned} & \text{length of SB} \\ =\; & (75\; \rm mi) \times \frac{\sin(40^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 55.12\; \rm mi\end{aligned}.

In other words, the distance between the northern lighthouse and the boat is approximately 30.73\; \rm mi, whereas the distance between the southern lighthouse and the boat is approximately 55.12\; \rm mi. Hence the conclusion.

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