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MrRa [10]
3 years ago
13

Porky the pig weighs 1,125 lbs. Less than Kobe the Cow. After each gains 50lbs. Kobe the Cow will weigh ten times as much as por

ky the pig. How much does each animal currently weigh?
Mathematics
1 answer:
Papessa [141]3 years ago
4 0

Answer:

Step-by-step explanation:

Let x represent the initial weight of Porky the pig.

Porky the pig weighs 1,125 lbs less than Kobe the Cow. This means that the initial weight of Kobe the cow is

x + 1125

After each gains 50lbs. Kobe the Cow will weigh ten times as much as porky the pig. This means that

(x + 1125) + 50 = 10(x + 50)

x + 1125 + 50 = 10x + 500

10x - x = 1125 + 50 - 500

9x = 675

x = 675/9

x = 75

Porky's current weight is

75 + 50 = 125 lbs

Kobe's current weight is

(75 + 1125) + 50

= 1250 lbs

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The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(
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Answer:

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

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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}.

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\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}.

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<h3>(d)</h3>

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