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3 years ago

13

A trapezoid was broken into a triangle and rectangle.

2 answers:

zloy xaker [14]3 years ago

6 0

Answer:

the area cm2

Step-by-step explanation:

777dan777 [17]3 years ago

4 0

Rectangle is 120
Triangle is 28
I think (don’t know for sure) but I think the trapezoid would just be the two added together so 148?? Maybe

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The graph shows f(x) = 1/2 and its translation, g(x).

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As you can see, each point on f(x) is moved up 4 units to get to g(x), so the function is g(x) = f(x) + 4. The f(x) function cannot possibly be f(x)=1/2, though, because that would be a horizontal line through y = 1/2 and that function is clearly not a horizontal line. So whatever f(x) is REALLY, add 4 to the tail end of it to show its translation.

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3 years ago

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Use the number line below, where RS equals 3 y plus 2 , ST equals 2 y plus 7 , and RT equals 10 y minus 11 .

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3y + 2 + 2y + 7 = 10y - 11
5y + 9 = 10y - 11
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4 years ago

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Sara is ordering t-shirts for her school. Sally's Shirt Shop charges $8.50 for each shirt and a one-time set up fee of $60. Shau

tensa zangetsu [6.8K]

Answer:

A ) 8.50x + 11 < 24x + 60

B ) 8.50x + 24 < 11x + 60

C ) 11x + 24 < 8.50x + 60

D ) 8.50x + 60 < 11x + 24

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.

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4 years ago

Is there anyone that can help me with a test?

Reptile [31]

Sure but where is the test?

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3 years ago