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Anon25 [30]
3 years ago
15

An individual has 3 different email accounts. Most of her messages, in fact 60% come into account #1, whereas 30%come into accou

nt #2, and the remaining 10% into account #3. Of the messages into account #1, only 1% are spam, whereas the corresponding percentages for account #2 and #3 are 3% and 5%, respectively. What is the probability that a randomly selected message is spam?
Mathematics
1 answer:
laila [671]3 years ago
7 0

Answer:

the probability is 0.02 (2%)

Step-by-step explanation:

defining the event S= the message is spam , then the probability is

P(S) = probability that a randomly selected message comes from account #1 * probability that the message is spam given that the message is from account #1 + probability that a randomly selected message comes from account #2 * probability that the message is spam given that the message is from account #2 + probability that a randomly selected message comes from account #3 * probability that the message is spam given that the message is from account #3 = 0.6 * 0.01 + 0.3 * 0.03 + 0.1 * 0.05 = 0.02 (2%)

thus the probability is 0.02 (2%)

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If x^2+y^2=1, what is the largest possible value of |x|+|y|?
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If <em>x</em>² + <em>y</em>² = 1, then <em>y</em> = ±√(1 - <em>x</em>²).

Let <em>f(x)</em> = |<em>x</em>| + |±√(1 - <em>x</em>²)| = |<em>x</em>| + √(1 - <em>x</em>²).

If <em>x</em> < 0, we have |<em>x</em>| = -<em>x</em> ; otherwise, if <em>x</em> ≥ 0, then |<em>x</em>| = <em>x</em>.

• Case 1: suppose <em>x</em> < 0. Then

<em>f(x)</em> = -<em>x</em> + √(1 - <em>x</em>²)

<em>f'(x)</em> = -1 - <em>x</em>/√(1 - <em>x</em>²) = 0   →   <em>x</em> = -1/√2   →   <em>y</em> = ±1/√2

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<em>f(x)</em> = <em>x</em> + √(1 - <em>x</em>²)

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