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Nesterboy [21]
1 year ago
15

There were some people on a train

Mathematics
2 answers:
Feliz [49]1 year ago
7 0
I think the answer is 31
quester [9]1 year ago
6 0
You minus 66 by 22 which is 44 than you add 17 which is 61 so the answer is 61
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Katrina owns a 9-acre farm. Last month, she bought her neighbor’s adjoining 8-acre farm. How large is Katrina’s farm now?
blondinia [14]

Answer:

17 acres

Step-by-step explanation:

9 acres + 8 acres = 17 acres, if that wasn't what the question meant please let me know and I'll change the answer or try to help clear confusion.

3 0
3 years ago
Joe's father is 15 more than twice Joe's age. How old is Joe's father? Solve algebraically.
dalvyx [7]

Answer:

Joe's father is 45.

Step-by-step explanation:

3x-x+2=4

6 0
3 years ago
Read 2 more answers
PLZ HELP I GIVE BRAINLIEST!!!
GREYUIT [131]

Answer:

800 × ( 3.87÷ 100 ) × 1

= 30.96

8 0
3 years ago
One condition for performing a hypothesis test is that the observations are independent. Marta is going to take a sample from a
AleksandrR [38]

Answer:

The correct answer to the following question will be "60 students".

Step-by-step explanation:

Marta will be taking a sampling frame from some kind of 600 student group.

Mean,

N = 60  

Although the sampling method could perhaps consist of the following components 10% of the population,

⇒  600\times 10 \ percent

⇒  60

In order to view these findings as autonomous, 60 students would then have to analyze Marta lacking replacements.

6 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
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