What is a good reasing 8x−12y?

A coin collection is made up of 34 coins, all nickels and dimes. The total value of the collection is $1.90. How many nickels ar

wel

Answer:

Hey There the Answer to this is x = nickels

y= dimes

x + y = 34

.05x + .10y= $1.90

-.10x - .10y = -3.40

-.05x + .10y = 1.90

-.05x = -1.50

x = 30 nickels

30 + y = 34

y = 4 dimes

(3,4) And that you get the answer.

Hope it helps!

6 0

4 years ago

Y=5

Ymorist [56]

15. = slope would be 7 and y-int would be -4
16. = slope would be -2/5 and y-int would be 0
17. = doesn’t have a y variable

slope intercept form is y=mx+b with the m being the slope and the b being the y-int. in some cases where the equation is not in this form you have to change it so it is in that form by using opposite operations

5 0

3 years ago

PLEASE HELP FAST The Booster Club raised $30,000 for a sports fund. No more money will be placed into the fund. Each year the fu

zloy xaker [14]

Answer:

c. $24,435

Step-by-step explanation:

$30000(1 - .05)^{4} = 24,435.19$

3 0

3 years ago

A rectangular prism and its dimensions are shown in the diagram.

lilavasa [31]

Answer:

51.08 in.²

Step-by-step explanation:

S.A = (l x b) + (l × h) + (b × h)

= (8.2 in. × 3.4 in.) + (8.2 in. × 2 in.) + (3.4 in. × 2 in.)

= 27.88 in.² + 16.4 in.² + 6.8 in.²

= 51.08 in.²

8 0

3 years ago

Read 2 more answers

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

4 years ago