1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
11111nata11111 [884]
3 years ago
15

Michelle wants to make cupcakes for her daughter's birthday. The recipe calls for 3/4 cup of brown sugar, 1 1/2 cups of white su

gar and 2 cups of powdered sugar and will make 12 cupcakes. how much sugar will be in each cupcake?
Mathematics
1 answer:
Oksanka [162]3 years ago
7 0
Well you want to add 3/4 , 1 1/2, and 2. Then divide that answer by 12.


You might be interested in
for a special concert by a popular band, the concert hall owner decided to charge $20 per ticket for the first "r" rows and $15
Alecsey [184]
R = 16 according to the problem, so the total amount of income will be:
$20*16 + $15*(40 - 16)
= $320 + <span>$15*(24)
= </span>$320 + <span>$360
= $680
that is the income, but the expenses are $275 so the total revenue is the subtraction of those two
total revenue = $680 - $275
= $405</span>
7 0
3 years ago
There is a bag filled with 5 blue and 6 red marbles.
Afina-wow [57]

Answer:

The probability of getting two of the same color is 61/121 or about 50.41%.

Step-by-step explanation:

The bag is filled with five blue marbles and six red marbles.

And we want to find the probability of getting two of the same color.

If we're getting two of the same color, this means that we are either getting Red - Red or Blue - Blue.

In other words, we can find the independent probability of each case and add the probabilities together*.

The probability of getting a red marble first is:

\displaystyle P\left(\text{Red}\right)=\frac{6}{11}

Since the marble is replaced, the probability of getting another red is: \displaystyle P\left(\text{Red, Red}\right)=\frac{6}{11}\cdot \frac{6}{11}=\frac{36}{121}

The probability of getting a blue marble first is:

\displaystyle P\left(\text{Blue}\right)=\frac{5}{11}

And the probability of getting another blue is:

\displaystyle P\left(\text{Blue, Blue}\right)=\frac{5}{11}\cdot \frac{5}{11}=\frac{25}{121}

So, the probability of getting two of the same color is:

\displaystyle P(\text{Same})=\frac{36}{121}+\frac{25}{121}=\frac{61}{121}\approx50.41\%

*Note:

We can only add the probabilities together because the event is mutually exclusive. That is, a red marble is a red marble and a blue marble is a blue marble: a marble cannot be both red and blue simultaneously.

4 0
3 years ago
Which equation can be used to find AB?
kogti [31]

Answer:

cos 42 = 10 / AB

Step-by-step explanation:

Since this is a right angle, we can use trig functions

cos theta = adjacent / hypotenuse

cos 42 = 10 / AB

5 0
3 years ago
Read 2 more answers
126.4% of what number is 158?
Maksim231197 [3]
<span>126.4% of 158 = 199.712</span>
6 0
3 years ago
PLEASEE HELPP. Im rlly struggling please help if you can. with steps
Nadusha1986 [10]

Answer:

lol

Step-by-step explanation:

L O LDJDDHDHDHDHFFHFVDVDVDBSSBDBDBDBD

7 0
1 year ago
Read 2 more answers
Other questions:
  • What is 14x -2y^2+38x-23y^2
    7·1 answer
  • Eight less than four times a number is less than 56. What are the possible values of that number
    12·2 answers
  • The area of a floor is 48 ft2. If 20% of the floor is covered with an area rug, hue much floor is space remains uncovered
    5·2 answers
  • Two friends share 1/4 gallon of lemonade equally.what fraction does each friend get
    13·1 answer
  • I need to know how to work<br> This
    10·1 answer
  • Find the area of the shaded region of the graph
    9·1 answer
  • I NEED HELP PLEASE, THANKS! :)
    12·1 answer
  • HELP ASAP. The two lines graphed on the grid each represent an equation.
    14·1 answer
  • The line with equation y = 2x - 6 passes through the point (a,2). The value of a is:
    9·1 answer
  • A C B, X Z Y
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!