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3 years ago

These paint cans come in different sizes. Let's try calculating the unit prices of each

Mathematics

1 answer:

Finger [1]3 years ago

6 0

Answer:

See Explanation

Step-by-step explanation:

The gallon sizes and prices are given below:

$\left|\begin{array}{c|c|C}Gallon\: Size&Price(\$)&\text{Unit Price Per Gallon}\\--------------&-----&---------\\4\text{(with 50\% Extra)}=6 \:Gallons&36&36\div 6=\$6 \:per\:gallon \\3&24&24\div 3=\$8 \:per\:gallon\\2 &20&20\div 2=\$10\:per\:gallon\end{array}\right|$

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PLEASE HELP it should be easy for someone but I can’t do it

-Dominant- [34]

Answer:

Total crackers on the plate are 12

Step-by-step explanation:

Manuel ate crackers = $\frac{1}{3}$

His brother ate crackers = $\frac{1}{4}$

Crackers left on the plate = 5

We need to find how many crackers were there on the plate.

Let x be the total crackers on the plate

So, we can write the equation

$x-\frac{1}{3}x-\frac{1}{4}x=5$

Because 1/3 and 1/4 crackers are eaten and 5 are left so, we subtract 1/3x and 1/4x from x and equal it to 5

$\frac{12x-4x-3x}{12}=5\\\frac{12x-7x}{12}=5\\\frac{5x}{12}=5\\x=\frac{5*12}{5}\\x=12$

So, we get x = 12

Therefore, total crackers on the plate are 12

8 0

3 years ago

Use the drop-down menus to identify the key values of the box plot. The median is . The minimum is . The maximum is . The lower

Elenna [48]

Answer:

median-c

minimum-a

maximum-e

Q1-b

Q2-d

6 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

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

Ay/b-by/a=a^2-b^2. find value of y

klemol [59]

We can actually undsitribute the y

$y( \frac{a}{b}- \frac{b}{a})=a^2-b^2$
divide both sides by $( \frac{a}{b}- \frac{b}{a})$
$y= \frac{a^2-b^2}{ \frac{a}{b}- \frac{b}{a} }$
times the right side by (ab/ab)
$y= \frac{ab(a^2-b^2)}{ a^2- b^2 }$
y=ab

4 0

3 years ago

What is 1/2+3/4= ...........

Mademuasel [1]

To answer the calculation 1/2 + 3/4, you need to do the following:

1/2 + 3/4 = 2/4 + 3/4 = (2 + 3) / 4 = 5/4

This can be also written as $1\frac{1}{4}$ .

3 0

3 years ago