All categories

3 years ago

U have 12 dogs all but 8 dogs died how many dogs are still alive

Mathematics

2 answers:

uysha [10]3 years ago

7 0

Hey you have to subtract the total dogs from the died dogs
which is 12 - 12 = 0
You have still no dogs alive.

mash [69]3 years ago

4 0

8 dogs are still alive because you take away 4 from 12 which equals 8
.

You might be interested in

A group of students surveyed chose baseball and soccer as their favorite sport in the ratio 3:8

Anna [14]

$3x-\ those\ who\ chose\ baseball\\8x-\ those\ who\ chose\ soccer\\\\
a)\\3x+8x=132\\\\
11x=132\ \ \ | divide\ by\ 11\\\\x=12\\\\
8x=8*12=96\\\\
96\ chose\ soccer.\\\\\
b)\\8x=56\ \ \ | divide\ by\ 8\\\\x=7\\\\
3x=3*7=21\\\\
21\ chose\ baseball.$

5 0

3 years ago

5pm to 5 am how many hours

slega [8]

Answer:

12 hours

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Compare one third to sixths

zhuklara [117]

In one third you would need 12 one thirds to make a whole sixth

6 0

3 years ago

Read 2 more answers

A car dealership pays $8,550 for a car. They markup the price by 20.4% to get the retail price. What is the retail price of the

marin [14]

Answer:

$10294.20

Step-by-step explanation:

8550 increase 20.4% =

8550 × (1 + 20.4%) = 8550 × (1 + 0.204) = 10294.2

3 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:

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.

4 0

3 years ago