A limited-edition poster increases in value each year with an initial value of $18. After 1 year and an increase of 15% per year
2 answers:
Answer:
The required equation is .
Step-by-step explanation:
Consider the provided information.
The Initial value of poster = $ 18
After 1 year amount of increase = $ 20.70
With the rate of 15% = 0.15
Let future value is y and the number of years be x.
Now verify this by substituting x=1 in above equation.
Which is true.
Hence, the required equation is .
You might be interested in
Answer:
a)
b)
c)
And we can find the usual limits with:
And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.
d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.
And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.
Step-by-step explanation:
We have the following data:
11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2
Part a
We can calculate the sample mean with the following formula:
And if we replace we got:
For the median we need to sort the values on increasing order and we have:
8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8
Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:
The mode would be the most repeated value on this case:
Part b
The range is defined as and if we replace we got:
We can calculate the standard deviation with the following formula:
And if we replace we got:
Part c
For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.
We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and
And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got
Then we can find the IQR like this:
And we can find the usual limits with:
And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.
Part d
The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.
And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.