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tangare [24]
3 years ago
10

Alicia shoots A basketball at a hoop hundred times. She hits the backboard and Misses with 2/5 of her shots, hit the rim and mis

ses with 32% of he shots, and make a basket with the rest of her shots. How many baskets does she make?
Mathematics
1 answer:
g100num [7]3 years ago
4 0

Answer:

28 baskets

Step-by-step explanation:

Total Shots = 100

Backboard and miss = 2/5 of total

That is:

\frac{2}{5}*100=40

Rim and miss = 32% of total

That is:

32% = 32/100 = 0.32

0.32 * 100 = 32 shots

So, miss = 40 + 32 = 72

The rest, she makes, so left:

100 - 72 = 28 shots

Alicia makes 28 baskets (from 100)

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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
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(c) 0.906

Step-by-step explanation:

Let's denote the events as follows:

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<em>S</em> = The email is spam

<em>V</em> = The email is valid.

The information provided to us are:

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        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.

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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:

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The probability that a message is valid provided that it does not contain free is approximately 0.906.

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