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poizon [28]
3 years ago
13

Is 1.875 and rational number

Mathematics
1 answer:
jonny [76]3 years ago
6 0

Answer: 1.875 is a rational number because it can be represented as a ratio or a fraction.

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Please help with at least one of these( I accept 1 answer if you have it; make sure you are 100% positive its correct though!)
AlladinOne [14]

Answer:

Step-by-step explanation:

7 0
3 years ago
An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in 60% of the spam messages and only 4% of
ANEK [815]

Answer:

(a) 0.152

(b) 0.789

(c) 0.906

Step-by-step explanation:

Let's denote the events as follows:

<em>F</em> = The word free occurs in an email

<em>S</em> = The email is spam

<em>V</em> = The email is valid.

The information provided to us are:

  • The probability of the word free occurring in a spam message is,             P(F|S)=0.60
  • The probability of the word free occurring in a valid message is,             P(F|V)=0.04
  • The probability of spam messages is,

        P(S)=0.20

First let's compute the probability of valid messages:

P (V) = 1 - P(S)\\=1-0.20\\=0.80

(a)

To compute the probability of messages that contains the word free use the rule of total probability.

The rule of total probability is:

P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})

The probability that a message contains the word free is:

P(F)=P(F|S)P(S)+P(F|V)P(V)\\=(0.60*0.20)+(0.04*0.80)\\=0.152\\

The probability of a message containing the word free is 0.152

(b)

To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

P(A|B)=\frac{P(B|A)P(A)}{P(B)}

The probability that a message is spam provided that it contains free is:

P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947\\

The probability that a message is spam provided that it contains free is approximately 0.789.

(c)

To compute the probability of messages that are valid given that they do not contain the word free use the Bayes' Theorem. That is,

P(A|B)=\frac{P(B|A)P(A)}{P(B)}

The probability that a message is valid provided that it does not contain free is:

P(V|F^{c})=\frac{P(F^{c}|V)P(V)}{P(F^{c})} \\=\frac{(1-P(F|V))P(V)}{1-P(F)}\\=\frac{(1-0.04)*0.80}{1-0.152} \\=0.90566

The probability that a message is valid provided that it does not contain free is approximately 0.906.

4 0
3 years ago
Six to the third power,
scoundrel [369]
6 to the third power would be 6x6x6 which equals 216. All you do to find the power is multiply the number by its self the amount of times it asks like 4 to the second power would be 16 and 8 to the 4 power would be 4096
4 0
3 years ago
Read 2 more answers
How many solutions does 73x-37=73x -37
bonufazy [111]

Answer:

0 is the answer.

73x -73x = -37+37

0=0

3 0
3 years ago
Solve for x. (10−5 2/3) ⋅ (−2)+x=8 2/3
Pepsi [2]

Answer:

17.3 or fraction 17 1/3

Step-by-step explanation:

7 0
3 years ago
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