Answer:
its the second one
Step-by-step explanation:
Answer:
Step-by-step explanation:
Keywords:
System of equations, variables, cost, tickets, adults, children.
For this case we must solve a system of equations with two variables represented by the tickets of students and adults of a school production.
We define the variables according to the given table:
a: Number of tickets sold to adults
c: Amount of tickets sold to children.
We then have the following system of equations:
A + c = 67
10a + 5c =440
From the first equation, we clear the value of the variable c:
C = 67 - a
Answer:
The value that could replace c in the table is:
C = 67 - a
Option C is the answer!
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Answer:

Step-by-step explanation:
Given



Required
Determine P(Gray and Blue)
Using probability formula;

Calculating P(Gray)



Calculating P(Gray)



Substitute these values on the given formula



1000000000 because less than this number is 9 digits and more than this numbet is not least
Answer: The relative frequency of column A in group 1=
The relative frequency of column B in group 1=
The relative frequency of column A in group 2=
The relative frequency of column B in group 2=
Step-by-step explanation:
The relative frequency is the ration of each frequency by the total value in particular group.
For group 1 : Total =102+34=136
The relative frequency of column A in group 1=
The relative frequency of column B in group 1=
For group 2 : Total =18+14=32
The relative frequency of column A in group 2=
The relative frequency of column B in group 2=