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bulgar [2K]
2 years ago
9

In the diagram below, what is the value of X?

Mathematics
1 answer:
Agata [3.3K]2 years ago
7 0

Answer:

x=15

Step-by-step explanation:

Well we start with the fundamentals of a triangle we know that all the angles of a triangle add up to 180*, so you can start by adding 55* and 65* and end up with 120* then you can add 30* and get 150* then you can take the total of 150* and subtract it from 180* to get 30 now you take that 30 and divide it by 2 and get 15* for the final angle of the triangle.

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\bf \cfrac{2}{5+i}\cdot \cfrac{5-i}{5-i}\implies \cfrac{2(5-i)}{\stackrel{\textit{difference of squares}}{(5+i)(5-i)}}\implies \cfrac{2(5-i)}{\stackrel{\textit{recall }i^2=-1}{5^2-i^2}}\implies \cfrac{2(5-i)}{25-(-1)} \\\\\\ \cfrac{2(5-i)}{25+1}\implies \cfrac{2(5-i)}{26}\implies \cfrac{5-i}{13}

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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}.

(c) \displaystyle P(B) = \frac{9}{100}.

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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:

\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}.

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