You can use calculus ( related rates). Given the rate os change of the radius for example you can find the rate of change of volume using differential calculus.
If you graphed the equation: y = 5
You would put a straight line going horizontally through 5. Like this picture shows. If it were x = 5, for example, then it would go up and down through 5.
Let me know if this helped! Have a nice day!
Answer:
a) Cost of the child-care center: y = 300 + 79x
Cost of the family-run childcare: y = 180 + 93x
Where x is the number of weeks.
b) In the short-term, the family-run childcare will be cheapest because the charge for enrolling is less than for the child-care center.
c)The long term option (child-care center) becomes more cost-effective if you plan to enroll your child for more than 8.6 weeks.
Step-by-step explanation:
Hi there!
a)Let y be the cost of childcare. In the child-care center, the cost is 300 plus 79 per week. Let x be the number of weeks, then the cost function for the child-care center will be:
y = 300 + 79x
In the same way, the cost of the family-run childcare will be:
y = 180 + 93x
b) In the short-term, the family-run childcare will be cheapest because the charge for enrolling is less than for the child-care center. Let´s calculate the cost of both childcare options for low values of x:
Child-care center:
y = 300 + 79 x
x = 0, y = 300
x = 1, y = 379
x = 2, y = 458
x = 3, y = 537
x = 4, y = 616
Family-run:
y = 180 + 93x
x = 0, y = 180
x = 1, y = 273
x = 2, y = 366
x = 3, y = 459
x = 4, y = 552
c) We have to find the value of x for which the cost of the family-run childcare is greater than the cost of the child-care center:
cost family-run > child-care center
180 + 93x > 300 + 79 x
93x -79x >300 - 180
14x > 120
x > 120/14
x > 8.6
The long term option (child-care center) becomes more cost-effective if you plan to enroll your child for more than 8.6 weeks.
Answer:
μ = 235.38
σ = 234.54
Step-by-step explanation:
Assuming the table is as follows:
![\left[\begin{array}{cc}Savings&Frequency\\\$0-\$199&339\\\$200-\$399&86\\\$400-\$599&55\\\$600-\$799&18\\\$800-\$999&11\\\$1000-\$1199&8\\\$1200-\$1399&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7DSavings%26Frequency%5C%5C%5C%240-%5C%24199%26339%5C%5C%5C%24200-%5C%24399%2686%5C%5C%5C%24400-%5C%24599%2655%5C%5C%5C%24600-%5C%24799%2618%5C%5C%5C%24800-%5C%24999%2611%5C%5C%5C%241000-%5C%241199%268%5C%5C%5C%241200-%5C%241399%263%5Cend%7Barray%7D%5Cright%5D)
This is an example of grouped data, where a range of values is given rather than a single data point. First, find the total frequency.
n = 339 + 86 + 55 + 18 + 11 + 8 + 3
n = 520
The mean is the expected value using the midpoints of each range.
μ = (339×100 + 86×300 + 55×500 + 18×700 + 11×900 + 8×1100 + 3×1300) / 520
μ = 122400 / 520
μ = 235.38
The variance is:
σ² = [(339×100² + 86×300² + 55×500² + 18×700² + 11×900² + 8×1100² + 3×1300²) − (520×235.38²)] / (520 − 1)
σ² = 55009.7
The standard deviation is:
σ = 234.54
Answer:
y= -1/2 x -1
Step-by-step explanation:
