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3 years ago

Perform the indicated operation.

Mathematics

2 answers:

kirza4 [7]3 years ago

7 0

Step-by-step explanation:

1st.

(7x+2×2-6x-4) - (2x+3×2-3x+11)

(7x+4-6x-4)-(2x+6-3x+11)

collect like terms

(7x-6x-4+4)-(2x-3x+11+6)

(x+0)-(-x+17)

multiply - by the bracket

x+0+x-17

collect like terms

x+x+0-17

2x-17=0

2x=17

divide both sides by the coefficient of x

2x/2=17/2

x=17/2

Ivanshal [37]3 years ago

5 0

Answer Solution:4x2 - 2x) - (-5x2 - 8x)

4x2 - 2x) - (-5x2 - 8x)= 4x2 - 2x + 5x2 + 8x.

4x2 - 2x) - (-5x2 - 8x)= 4x2 - 2x + 5x2 + 8x.= 4x2 + 5x2 - 2x + 8x.

4x2 - 2x) - (-5x2 - 8x)= 4x2 - 2x + 5x2 + 8x.= 4x2 + 5x2 - 2x + 8x.= 9x2 + 6x.

but this is the answer = 3x(3x + 2).

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3 years ago

A furniture designer builds a trapezoidal desk with a semicircular cutout. What is the area of the desk?

nirvana33 [79]

Answer:

$Area = 26478cm^2$

Step-by-step explanation:

Given

See attachment for desk

Required

The area

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2 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

Step-by-step explanation:

Let's denote the events as follows:

F = The word free occurs in an email

S = The email is spam

V = 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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3 years ago