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3 years ago

15

What is the length of AB

1 answer:

sweet [91]3 years ago

6 0

Answer:

6 units is the answer im pretty sure

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May i have help with this

Leya [2.2K]

Answer: D. 7^11

Step-by-step explanation:

For exponents with the same base, we just add the exponents and keep the same base. 7 is the same base; therefore 5+6=11 which means the answer is 7^11

3 0

2 years ago

Which MATH is a parallelogram.<br><br> A.<br> 3<br> B.<br> 4<br> C.<br> 7<br> D.<br> 12

joja [24]

A parallelogram have 4 sides

4 0

3 years ago

Determine the intervals on which the function is (a) increasing and (b) decreasing

Sav [38]

These intervals are measured by the x-values. Additionally, intervals of increase and decrease are always express with parentheses in interval notation. Therefore, the decreasing interval is (-2, 0)U(1, 3) and the increasing interval is (0, 1).

Hope this helps!

5 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

(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

4 years ago

7(3.5+2)+8(9.3-7) please help

CaHeK987 [17]

Answer:

awdwdadadawdawdawdw

Step-by-step explanation:

awdadwdawddwadawdwa

8 0

3 years ago

Read 2 more answers