Using it's concept, it is found that there is a 0.0366 = 3.66% probability that your coach and your friend get orange and you get a fruit-punch.
<h3>What is a probability?</h3>
A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.
In this problem, there are 15 bottles.
- 5 are orange, hence the is a 5/15 = 1/3 probability that the coach gets orange, hence P(A) = 1/3.
- After the coach, there will be 14 bottles remaining, of which 4 are orange, hence the probability that the friend gets orange is of P(B) = 4/14 = 2/7.
- For you, there will be 13 bottles remaining, of which 5 will be of fruit-punch, hence the probability you get fruit-punch is of P(C) = 5/13.
The probability of the three outcomes occurring is given by:
0.0366 = 3.66% probability that your coach and your friend get orange and you get a fruit-punch.
More can be learned about probabilities at brainly.com/question/14398287
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Answer:
Let P represent the number of pies sold and J represent the number of juices sold.
a)
The first equation for the sum of the sales:
P + J = 79
P = 79 - J
Next, for the sum of the amounts earned:
1.65P + 1.36J = 118.17
Substituting P,
1.65 (79 - J) + 1.36J = 118.17
J = 42
Using this value of J,
P = 79 - 42
P = 37
b)
Mr. Sanchez's class earned: 1.65 x 37 = $61.05
Mr. Kelly's class earned: 1.36 x 42 = $57.12
So Mr. Sanchez's class earned more money.
c) Mr. Sanchez's class earned:
61.05 - 57.12
= $3.93 more than Mr. Kelly's class
Step-by-step explanation:
:)
Answer:
The area of a label is 
Step-by-step explanation:
we know that
The lateral area of a cylinder (label of the can) is equal to
we have that
A can is 4 inches wide
so
The diameter of the can is 4 inches
----> the radius is half the diameter
----> height of the label
substitute in the formula
Steps to solve:
15 + 2a = 35
~Subtract 15 to both sides
2a = 20
~Divide 2 to both sides
a = 10
Best of Luck!
Answer:
k = 4
Step-by-step explanation:
16k = 16(4) = 64 and
64 = 4 × 4 × 4
![\sqrt[3]{64}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B64%7D)
= ![\sqrt[3]{4^{3} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B4%5E%7B3%7D%20%7D)
= 4
