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n200080 [17]
3 years ago
14

Rob notices that 55 percent of the people leaving the supermarket chhose plastic bags instead of paper bags. out of 600 people,

how many can rob predict will carry plastic bags?
Mathematics
1 answer:
Temka [501]3 years ago
4 0
To solve this answer, you are trying to find 55% of 600, because that is the number of people who chose plastic bag.

55% can be re-written as 55/100 OR 0.55 since percentages are always out of 100.

So, we do 0.55*600, which gives us 330. 330 people carry plastic bags.
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684 =<br> A<br> 600 + 80 + 40<br> B<br> 40 + 600 + 8<br> 6 + 80 + 400<br> D<br> 600 + 80 + 4
borishaifa [10]

Answer:

The answer is d. Hope this helps :D

Please mark brainliest!

4 0
2 years ago
Read 2 more answers
Write the slope intercept form of the equation of the line through the given points (-5,3) and (3,0)
grigory [225]

Answer:Y=8,-5

Step-by-step explanation:

3 0
3 years ago
What is quadratic equaton?
Lera25 [3.4K]
A quadratic equation is an equation that includes a squared term. For example, 3x + 7 = 28 is not a quadratic equation as it only has x, whereas x^2 + 5x + 6 = 0 is a quadratic equation as it includes x^2. Quadratic equations also usually have two solutions, whereas linear equations (like my first example) only have one.

I hope this helps! Let me know if you would like me to explain anything more :)
6 0
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

(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
The temperature was t degrees farenheight . It fell 8 degrees farenheight and is now 32 degrees farenheight.What was the orginal
LenKa [72]
Well since you subtract 8 from the original and now you have 32 you should add 8 which means the answer is 40 degrees Fahrenheit. tell me if i’m wrong
4 0
3 years ago
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