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1 year ago

6

ali goes for cycle ride. he starts at 3pm . he finished at 5 38pm .he has a total of 25 minutes rest during the ride. how long h

e spends cycling?

1 answer:

nordsb [41]1 year ago

4 0

Answer:

two hours and thirteen minutes.

Step-by-step explanation:

count up from 3pm to 5pm to start with. That gives you two hours

we already have 38 minutes, now just subtract the 25 minutes he spent resting.

That leaves you with 2 hours and 13 minutes.

Correct me if I'm wrong I'm currently responding to this at a major lack of sleep lol

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Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

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Step-by-step explanation:

<h3>(a)</h3>

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

<h3>(b)</h3>

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

<h3>(c)</h3>

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

<h3>(d)</h3>

By Bayes' Theorem:

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Similarly:

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They'll reach the same population in approximately 113.24 years.

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