Help please 8. What is mZHTA?

Write 10 x 10 x 10 x 10 x 10 with an exponent. 10 x 10 x 10 x 10 x 10 =

lakkis [162]

Answer:

$10^5$

Step-by-step explanation:

Exponents are basically just saying how many times the number is multiplied by itself.

In the expression $10\cdot10\cdot10\cdot10\cdot10$ , we can see that 10 is being multiplied by itself 5 times.

So the exponent becomes 5, and the base is 10.

$10^5$ .

Hope this helped!

7 0

3 years ago

I GIVEEE BRAINLILSTT

alekssr [168]

Answer:

(x, y) --> (x + 14, y + 8)

Step-by-step explanation:

Look at 1 original point and its corresponding translated point.

Let's look at F and F'.

To go from F to F', you need to go right in x 14 units.

Then you need to go up in y 8 units.

The translation rule is to add 14 to x and 8 to y.

(x, y) --> (x + 14, y + 8)

4 0

2 years ago

Read 2 more answers

Which are solutions to the following equation? x^2=21

creativ13 [48]

Answer:

x

=

±

√

21

Explanation:

To find

x

2

you have to find the square root of

21 x = ± √ 21 ≈ ± 4.5286

Step-by-step explanation:

4 0

2 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:

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

John read the first 114 pages of a novel, which was 3 pages less than start fraction, 1/3 1, divided by, 3, end fraction of the

anyanavicka [17]

HERE IS YOUR ANSWER

114 = (1/3)p - 3
HOPED I HELPED AND SMASH THAT THANKS BUTTON AND ADD ME AS A FRIEND

3 0

2 years ago