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zepelin [54]
3 years ago
5

-7-5+3=1 ? true or false

Mathematics
2 answers:
Lelechka [254]3 years ago
4 0
False would be the answer
lubasha [3.4K]3 years ago
4 0
When you add the negatives;
-7 and -5 you get -12.
-12 + 3 = -9
-9 doesn’t equal 1
So FALSE
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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
4 years ago
64.3 - 3 times 8 divided by 2<br> evaluate the expression
gavmur [86]

Answer:

245.2

Step-by-step explanation:

(64.3 - 3)8 / 2

(61.3)(8) / 2

490.4 / 2

= 245.2

6 0
3 years ago
What two digit number when multiplied by itself has a product of 625?​
In-s [12.5K]

Answer:

25*25

Step-by-step explanation:

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3 years ago
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A pet store currently has a total of 45 cats and dogs. There are 7 more cats than dogs. Determine the number of cats and dogs in
irina1246 [14]

x+y=45

x=y+7

Use substitution...

Plug y+7 into the first equation.  You get (y+7)+y=45.  Next combine the like terms.  2y+7=45.  Subtract 7 from both sides and you get 38.  Divide 2 on both sides to get y alone and you get 19.  Since there are more cats than dogs, this is the amount of dogs.  Add 19+7 since there are 7 more cats than dogs and you get 26 cats.  Therefore there is 19 dogs and 26 cats.

5 0
3 years ago
What is the area of this figure?
Mazyrski [523]
I believe the area is 66
8 0
3 years ago
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