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3 years ago

ellie brought 60 cookies to class today. 30% of them have mint filing. how many cookies have the mint filling

Mathematics

2 answers:

Pepsi [2]3 years ago

8 0

60 cookies X 0.30 = 18 mint filling

Elan Coil [88]3 years ago

5 0

So simple.

18.

If you need help to find why, here's a link

http://www.geteasysolution.com/what-is-30-percent-of-60

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A rectangle has a perimeter of 56 in. The length is four more than the width. What are the length and width of the rectangle

Aleks04 [339]

Answer:

Width = 12 in

Length = 16 in

Step-by-step explanation:

Let, the Width of the rectangle = w in

Now, the length of the rectangle = (w + 4) in

Now, Perimeter = 56 in

Also, we know that

Perimeter of the Rectangle = 2 (Length +Width)

or, 2 (Length +Width) = 56

⇒ 2(w + 4 +w) = 56

or, 4w =56 -8

width = 12 in

So, Length = w + 4 = 12 + 4 = 16 in

7 0

3 years ago

Yea uh... mmmmm... may someone help me with this. Thank you so very much! :D

kobusy [5.1K]

What’s the question?

5 0

3 years ago

Someone please help with this problem

den301095 [7]

A) write an equation you can solve to find the cost of one ticket
x=cost of one ticket.
9x+5=104

b)
I have to solve this equation.
1)adding -5 to each side
2) dividing both sides by 9

c)
9x+5=104
1) 9x+5-5=104-5
9x=99
2) 9x/9=99/9
x=11

the cost of each ticket was $11

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

There are four cars entered into a pinewood derby race. in how many different orders can they cross the finish line, assuming th

artcher [175]

This is a problem of fundamental counting principle. In a sequence of events, the total possible number of ways all events can performed is the product of the possible number of ways each individual event can be performed. So here there is one car in each place so there 4 possible cars to get the 1st place, 3 possible car for the 2nd place, 2 possible cars for 3rd place and 1 for the 4th place. The 4x3x2x1 = 24 different order

4 0

3 years ago