Warren has a container which holds 22 books of the same shape and size. The exact breakdown of his books is shown below:

Mathematics

1 answer:

ohaa [14]3 years ago

7 0

Given:

Total number of book = 22

Number of white books = 6

Number of Purple books = 9

Number of green books = 7

To find:

The probability that Warren takes out a green book in both draws (with replacement).

Solution:

We know that,

$\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$

$P(\text{White})=\dfrac{6}{22}$

$P(\text{Purple})=\dfrac{9}{22}$

$P(\text{green})=\dfrac{7}{22}$

After replacement, the probability of event remains the same and the independent.

The probability that Warren takes out a green book in both draws is:

$P=P(\text{green})\times P(\text{green})$

$P=\dfrac{7}{22}\times \dfrac{7}{22}$

$P=\dfrac{49}{484}$

Therefore, the correct option is (a).

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A small plane and a large plane are 6.8km from each other, at the same altitude (height). From an observation tower, the two air

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Answer:

7.9km

Step-by-step explanation:

(a)See attached for the diagram representing this situation.

(b)

In Triangle ABC

$\text{Using Law of Sines}\\\dfrac{\sin A}{a}=\dfrac{\sin C}{c} \\\dfrac{\sin A}{5.2}=\dfrac{\sin 58^\circ}{6.8} \\\sin A=5.2 \times \dfrac{\sin 58^\circ}{6.8}\\A=\arcsin (5.2 \times \dfrac{\sin 58^\circ}{6.8})\\A=40.43^\circ$

Next, we determine the value of Angle B.

$\angle A+\angle B+\angle C=180^\circ\\40.43+58+\angle B=180^\circ\\\angle B=180^\circ-(40.43+58)\\\angle B=81.57^\circ$

Finally, we find b.

$\text{Using Law of SInes}\\\dfrac{b}{\sin B}=\dfrac{c}{\sin C} \\\dfrac{b}{\sin 81.57^\circ}=\dfrac{6.8}{\sin 58^\circ} \\b=\dfrac{6.8}{\sin 58^\circ} \times \sin 81.57^\circ\\b=7.9km $ (to the nearest tenth of a kilometer)$