F(9) if f(x) = 5x

Four friends are planning to play 18 holes of golf. two of them need to rent clubs at $5 per set. total cart rental is $10. the

natka813 [3]

B how much did the friends pay in green fees

4 0

3 years ago

In the figure, m ll n. If the measure of angle 8 is (4x +7) and the measure of angle 2 is 107 degrees, what is the value of x? E

Archy [21]

Given:

$\begin{gathered} m\angle8=4x+7 \\ m\angle2=107 \end{gathered}$ $\begin{gathered} 4x+7+107=180\degree \\ 4x+114=180 \\ 4x=180-114 \\ 4x=66 \\ x=16.5 \end{gathered}$

3 0

1 year ago

The function below models the voltage, in volts, of a certain alternating current after x seconds, where A and b are positive co

Katarina [22]

First of all, we can just ignore A, it has no effect but to vertically stretch our cosine.

If it was only $f(x)=cosx$ , the function would be invertible as long as it's confined between $0$ and $\pi$ . Now, the argument of our cosine is not $x$ but $bx$ . It means that it won't stop at $\pi$ , but at $\frac{\pi}{b}$ .

Another way to think about it, "what should i replace x with so I get $\pi$ inside the cosine?"

4 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:

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

The manufacturer of a laser printer reports the mean number of pages a cartridge will print before it needs replacing is 12,425.

snow_lady [41]

A suitable probability app shows the 1st percentile to be at about 11040 pages.

_____

The app shown has an error in the first decimal digit. The number is closer to 11040.82301. We expect the manufacturer would round to the nearest 10 or 50 or 100 in advertising.

8 0

3 years ago

F(9) if f(x) = 5x - 16