The question is incomplete:
The dot plot shows the number of hours Stan worked each week for the last 3 months.
What fraction of the total number of weeks did Stan work 15 or more hours? Express in simplest form.
A. 5/7
B. 1/6
C. 1/5
D. 5/12
The image with the dot plot is attached.
Answer:
D. 5/12
Step-by-step explanation:
The dot plot shows the number of hours Stan worked and in order to know the number of weeks that he worked 15 hours or more, you have to count the dots that are between 15 and 18 hours. So, you have 3 dots in 15 hours, 1 dot in 16 hours and 1 dot in 18 hours. If you add this numbers (3+1+1=5), this means that he worked 5 weeks 15 or more hours. This number would be the numerator in your fraction. Then, the denominator would be the total amount of weeks and to find that you can count all the dots. The total number of dots is 12.
So, the fraction that represents the total number of weeks that Stan worked 15 or more hours is: 5/12. This is the simplest form because this numbers don't have any common factors other than 1.
you'd get a decimal value for the rate, just multiply it times 100, to get the percentage format of it.
Suppose we want to know the total cost of buying x toys and we know that each toy costs $2. The relationship between the cost and the number of toys is
C(x) = 2x
If we purchase 6 toys, the cost would be
C(6) = 2*6 = $12
This is an example where it's adequate to use a single variable (or unknown) to find the value of another variable.
Now suppose we want to know the total cost of buying x toys for $2 each and include the tax rate in the calculations.
If we know the tax rate r, we can compute the total cost as
C(X,r) = 2x*(1 + r/100)
For example, to purchase x=6 toys and the tax rate is r=8%, the total cost is:
C(6,8) = 2*6*(1 + 8/100)=$12.96
If we had tried to calculate this cost without the use of two unknowns, it would have not been possible.
Thus, the pattern to use one or two variables depends on how many factors determine the final result.
I take it you meant above to be a multiplication, so then,