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Sergio039 [100]
3 years ago
5

Jon's grades from reading class are shown below.

Mathematics
1 answer:
Paul [167]3 years ago
8 0

Answer:

28

Step-by-step explanation:

range=highest value-lowest value

range=100-72

range=28

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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
3 years ago
Li believes that the graph shows a direct variation. Why is Li incorrect in saying that the graph shows a direct variation?
padilas [110]

Answer:

Option 2) When the x-value is 0, the y-value is 1.

Step-by-step explanation:

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form y/x=k or y=kx

In a proportional relationship the constant of proportionality k is equal to the slope m of the line <em>and the line passes through the origin </em>

Remember that in a direct variation

For x=0, the value of y is equal to zero too

therefore

The graph is not a direct variation , because

When the x-value is 0, the y-value is 1.

3 0
3 years ago
Read 2 more answers
I’m confused plz help!!
Xelga [282]

The explanation is on the picture I attached!

Hope you get it!

3 0
4 years ago
Someone Help <br><br><br>I’m in 6th grade
ad-work [718]

Answer:

2120

Step-by-step explanation:

8 0
3 years ago
What is the measure of angle 1
creativ13 [48]

Answer:

86

Step-by-step explanation:

if u add the angle #'s on the INSIDE, itll give 94, since a triangle isnt obtuse, it would have to add up to 180, subtract 94 from 180 and u get the answer :)

hope this helps

8 0
4 years ago
Read 2 more answers
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