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IrinaVladis [17]
3 years ago
8

80=-8a please I need the answer

Mathematics
1 answer:
Free_Kalibri [48]3 years ago
6 0
80=-8a 
We are trying to solve for a. So we need to get a alone. To do that we have to divide each side by -8. 
80/-8= -8a/-8
Once you have solved each side you will get the following. 
-10=a
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Andrej [43]
Hi

I belive you add the top side (4) plus the bottom side (9.5) witch would equal 13.5 times the hieght (3) equals 40.5 divided by 2 equals 20.25. the area is 20.25ft squared

another way to do it
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3 years ago
Find the probabilities for a standard normal random variable Z.
vampirchik [111]

Answer:

Step-by-step explanation:

find the attachment showing std normal curve symmetrical about y axis.

Equal probabilities on either side of the mean thus the total probability to the right of mean is 0.50

From the table we can find that

a) P(Z>2.5) = 0.5- area lying between 0 and 2.5

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b) P(1.2<z<2.2) = F(2.2)-F(1.2)

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The perimeter of a semicircle is 10.28 kilometers. what is the semicircle's area
alexgriva [62]
The answer is 6.27. Hope this helps.
8 0
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a line passes through point (-2, 5) and has the same y-intercept as line y = 3x-4. what is the equation for the line?
IrinaK [193]

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Step-by-step explanation:

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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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