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frozen [14]
3 years ago
5

What percent of 160 is 56?use the percent bar model.

Mathematics
1 answer:
attashe74 [19]3 years ago
6 0

Answer:

56 is 35% of 160

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equivalent = 30w 30(w) 30*w

the rest are non equivalent

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More brinliest if its right :D im horrible at math
olga55 [171]

Answer:

No

Step-by-step explanation:

because the unit rate is 6 miles per hour for the first one and 10 miles per hour for the second

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Solve for x -<br><br><img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%2B%205x%20%2B%206%20%3D%200%20%5C%5C%20%5C%5C%20" id="T
Volgvan

Hello,

x² + 5x + 6 = 0

a = 1 ; b = 5 ; c = 6

∆ = b² - 4ac = 5² - 4 × 1 × 6 = 25 - 24 = 1 > 0

2 solutions :

x_{1} =  \frac{ - b -  \sqrt{\Delta} }{2a}  =  \frac{ - 5 - 1}{2 \times 1}  =  \frac{ - 6}{2}  =  - 3

x_{2} =  \frac{ - b  +   \sqrt{\Delta} }{2a}  =  \frac{ - 5  + 1}{2 \times 1}  =  \frac{ - 4}{2}  =  - 2

S = { -3 ; -2}

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1 year ago
Find the missing angle measurements. Screenshot included.
zheka24 [161]
One of the missing angles is 45degrees 64+71=135 then subtract from 180 since it is a triangle. Then 64 are vertical angles with the one across from it which stays the same. The last angle would be 64+56=120 and 180-120=60 which would be the last angle. Hope this Helps... Juan
3 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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