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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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Which of the following statements says that a number is between -3 and 3?
Shalnov [3]
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B, </span><span>|x| < 3, would be your answer.
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3 years ago
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Situation:
omeli [17]

The half life of the substance is 6.92 days.

<h3>What is the Half life  ?</h3>

Half life can be defined as the time required to reduce the concentration to half of its initial value.

Here the equation given is

\rm N = N_o e^{-kt}

The concentration will reduce to half therefore

\rm   N= \dfrac{N_o}{2}

so,

\rm 39/2 = 39  e^{- 0.1001 * t_{1/2}}

ln 0.5 = - 0.1001 * t(1/2)

-0.693 = -.1001 * t(1/2)

t(1/2) = 6.92 days.

Therefore the half life of the substance is 6.92 days.

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brainly.com/question/24710827

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2 years ago
An instructor gives her class the choice to do 7 questions out of the 10 on an exam.
Maksim231197 [3]

Answer:

(a) 120 choices

(b) 110 choices

Step-by-step explanation:

The number of ways in which we can select k element from a group n elements is given by:

nCk=\frac{n!}{k!(n-k)!}

So, the number of ways in which a student can select the 7 questions from the 10 questions is calculated as:

10C7=\frac{10!}{7!(10-7)!}=120

Then each student have 120 possible choices.

On the other hand, if a student must answer at least 3 of the first 5 questions, we have the following cases:

1. A student select 3 questions from the first 5 questions and 4 questions from the last 5 questions. It means that the number of choices is given by:

(5C3)(5C4)=\frac{5!}{3!(5-3)!}*\frac{5!}{4!(5-4)!}=50

2. A student select 4 questions from the first 5 questions and 3 questions from the last 5 questions. It means that the number of choices is given by:

(5C4)(5C3)=\frac{5!}{4!(5-4)!}*\frac{5!}{3!(5-3)!}=50

3. A student select 5 questions from the first 5 questions and 2 questions from the last 5 questions. It means that the number of choices is given by:

(5C5)(5C2)=\frac{5!}{5!(5-5)!}*\frac{5!}{2!(5-2)!}=10

So, if a student must answer at least 3 of the first 5 questions, he/she have 110 choices. It is calculated as:

50 + 50 + 10 = 110

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2 and a half hours estimate
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