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Volgvan
3 years ago
13

Can someone help me with this

Mathematics
1 answer:
insens350 [35]3 years ago
7 0
Let's solve this problem step-by-step.

STEP-BY-STEP EXPLANATION:

We will be using simultaneous equations to solve this problem.

The sum of angles on a straight line is 180°.

( R ) and ( 2x + 5 ) are both on the same straight line.

Therefore:

Equation No. 1 -

R + 2x + 5 = 180

R = 180 - 2x - 5

R = 175 - 2x

Vertically opposite angles are equivalent to each other.

( R ) is vertically opposite ( 3x + 15 ).

Therefore:

Equation No. 2 -

R = 3x + 15

Substitute the value of ( R ) from the first equation into the second equation to solve for ( x )

R = 3x + 15

175 - 2x = 3x + 15

- 2x - 3x = 15 - 175

- 5x = - 160

x = - 160 / - 5

x = 160 / 5

x = 32

Next we will substitute the value of ( x ) from the second equation into the first equation to solve for ( R ).

Equation No. 2 -

R = 175 - 2x

R = 175 - 2 ( 32 )

R = 175 - 64

R = 111

FINAL ANSWER:

Therefore, the answer is:

R = 111

x = 32

Hope this helps! :)
Have a lovely day! <3
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Olegator [25]
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A gift box’s rectangular base has a perimeter of 92 centimeters. The length of the base is one more than twice the base’s width.
kozerog [31]
Assume that the length of the rectangle is "l" and that the width is "w".

We are given that: 
(1) The length is one more than twice the base. This means that:
l = 2w + 1 .......> equation I
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92 = 2(l+w) ...........> equation II

Substitute with equation I in equation II to get the width as follows:
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3 0
3 years ago
A right angle has a league of 13 cm in the hypotenuse of 21 cm what is the length of the other leg
natta225 [31]
You can use c^2 = a^2 + b^2 to solve for the length of the other leg. Just substitute the values to the variable then transpose to get the value of the missing variable.

Given: a = 13; c = 21
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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
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