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3 years ago

How many fifths are in 6 4\5? Use your answer to rewrite 6 4\5 in the form \5?

1 answer:

7 0

6 4/5 is 34/5. This is because you take the 6 and multiply it by 5, the denominator. Next, you add four to that quantity, and get 34. So, to answer your question, there are 34 fifths in 6 4/5 and it can be written as 34/35

Hope this helped!!! :)

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Find m angle dbc in the figur below

Gemiola [76]

Answer:

It like the once like Line with smaller angle (DBC) measure wouldn't 30* and by the exam the ratio of angle use by subjective by multiplying as large of angle.

oh, by the way I think I hope I'm right so good luck with that?

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3 years ago

Mix number for 8x1/2

Paul [167]

Answer:

Step-by-step explanation:

Easy way to do this is;

8 _

8 * 2 = 16

16 + 1 = 17

17/2 = 8 , 1/2

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3 years ago

Please help! I will give brainlest answer

Sidana [21]

Connect the bdc letters together.

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3 years ago

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

Step-by-step explanation:

Let's denote the events as follows:

F = The word free occurs in an email

S = The email is spam

V = 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

What is the length, in meters, of the ramp, line segment AC, below? Round to the nearest whole number if necessary.

Brrunno [24]

10 meters.

Step-by-step explanation:

To find the missing side of a right triangle, we can use the Pythagorean theorem.

The Pythagorean theorem is...

$a^2+b^2=c^2$

Please be sure to note that a and b will always be the length of the sides and c will always be the hypotenuse, or in this case the ramp.

Since we know the values of both of the sides we can plug this into the theorem.

$6^2+8^2=c^2$

Now we can solve.

$36+64+c^2$

Add.

$100=c^2$

Many people will make the mistake of thinking 100 is the final answer, but since c is squared we need to find the square root of 100.

$\sqrt{100} =10$

c = 10

The ramp is 10 meters.

6 0

3 years ago