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Ilya [14]
3 years ago
10

PLS HELP WILL MARK BRAINLIEST!

Mathematics
2 answers:
sukhopar [10]3 years ago
8 0

Answer:  Approximately 21.667 km

This value as a fraction is 65/3, but I think the decimal form is probably more practical in real world settings.

======================================================

Work Shown:

20 min = 1/3 of an hour, since 20/60 = 1/3. In other words, there are three chunks of 20 minutes that make up a full hour.

distance = rate*time

distance = (65 kph)*(1/3 hr)

distance = (65*1/3) km

distance = (65/3) km

distance = 21.667 km which is approximate

tigry1 [53]3 years ago
7 0

Answer:

Distance rate=Rate x Time

Distance= (75 km/h) 20 mins

Convert 20 mins in hours= 1/3 h

Distance= (75 km/h) (1/3 h)= 25 km.....

listen ok

Step-by-step explanation:

hope it help

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I'd like to assume that the triangle is a right triangle and you solve for x so...

133 + c = 180   (supplementary property to find one angle of triangle)
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Select the correct answer. Why is it important to budget some money for entertainment? A. It is easier to stick to a budget if y
IRISSAK [1]

Answer:

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Step-by-step explanation:

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How to comvert 5/8 into a decimal​
Maksim231197 [3]

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Read 2 more answers
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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andriy [413]

Answer:

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Step-by-step explanation:

SA = area of base + 1/2(perimeter of base)(slant height)

SA = 16 + 1/2(16)(5)

SA = 16 + 8(5)

SA = 16 + 40

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