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3 years ago

If ABC is reflecting across the y-axis ,what are the coordinates of c’?

Mathematics

1 answer:

Brums [2.3K]3 years ago

7 0

<h3>Answer: (-6, 3)</h3>

Explanation:

First step is to locate point C. It is at location (6, 3)

Reflecting this over the y axis will change the sign of the x coordinate, while keeping the y coordinate the same. We go from (6, 3) to (-6, 3)

The algebraic rule for this is $(x, y) \to (-x,y)$

Therefore, point C' is located at (-6, 3)

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Write an expression for a number that is 5 times greater than 2

ElenaW [278]

Answer:

x = 5x > 2

or just 5x > 2 ya its that

Step-by-step explanation:

4 0

3 years ago

The sum of 6 and the product of 3 and d

Novosadov [1.4K]

Answer:

6 + 3d

Step-by-step explanation:

First, we can translate from English words to mathematical operations:

– Product: the result of multiplication

– Sum: the result of addition

So, we can rephrase the original sentence as “the result of adding 6 and (the result of multiplying 3 and d)”

When we multiply a constant like 3 by a variable like d, we usually write the two next to each other, which would be “3d” in this case. We can replace that last sentence in symbols with 6 + 3d

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2 years ago

NEED HELP ASAP ! PLEASE

solong [7]

Answer:

C'D'=10

Step-by-step explanation:

CB=5, and C'B'=12.5

5*2.5=12.5

4*2.5=10

8 0

2 years ago

The image of (-9,-4) after a reflection over the y-axis?

lidiya [134]

Answer:

The answer is (-4,9)

Step-by-step explanation:

When you reflect across y=-x the x and y coordinates switch places and change signs.

7 0

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

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