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3 years ago

8

Given the event "a die lands with a 6 on top", which of the following is the complement of this event? Choose the correct ans

wer below.
A. The number of times the die lands with a 6 on top in n throws of the die
B. The die lands with a 3 on the top
C. The die lands with a 6 on the bottom
D. The die lands with a 1, 2, 3, 4, or 5 on top

1 answer:

Veronika [31]3 years ago

8 0

Answer:

The die lands with a 1, 2, 3, 4, or 5 on top ⇒ answer D

Step-by-step explanation:

- Complementary events are two outcomes of an event that are the

only two possible outcomes

- The Complement Rule states that the sum of the probabilities of

an event and its complement must equal 1

- P(A) + P(A') = 1

- Lets solve the problem

- The event A is:

" a die lands with a 6 on top "

∴ P(A) = $\frac{1}{6}$

∵ P(A) + P(A') = 1

∴ $\frac{1}{6}$ + P(A') = 1

- Subtract $\frac{1}{6}$ from both sides

∴ P(A') = $\frac{5}{6}$

∵ The die has six numbers from 1 to 6

∴ The other numbers than 6 are 1 , 2 , 3 , 4 and 5

∴ The complement of event A is:

" A die lands with 1 , 2 , 3 , 4 or 5 "

* The complement of the event is:

The die lands with a 1, 2, 3, 4, or 5 on top

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

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

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

(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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