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Ganezh [65]
3 years ago
12

3. Choose the best statistical question about lizards. O How many lizards do you have in your backyard? О What are the lengths o

f the lizards in your backyard? O How many bugs did a lizard eat today?​
Mathematics
1 answer:
ASHA 777 [7]3 years ago
4 0

Answer: what are lengths of lizard

Step-by-step explanation: you can calculate average length of lizard

And also standard deviation and determine median too

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3.5

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6x+18

Step-by-step explanation:

Multiply 3 to all the numbers on the parenthesis

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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
Graph the circle x2 + y2 – 14x – 12y + 76 = 0.
nikklg [1K]

Answer:

The answer is, the center is (7,6) and the radius 3 soo, you would count three from the center to the arc( outer line and then if its three than ur good. proven in khan academy don't worry

Step-by-step explanation:

XD finally no more schoooooolllllllll baaaabbyyyyyy

7 0
3 years ago
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