The condo costs $163,000, earns $2,986 per month, spends no more than 25% of her income, then if she pays $33,000 for the down payment, the remaining amount would be $130,000. Since 20% of the initial cost is only $32,600, she can adjust her down payment to 20.25% of the initial cost so that the annual payments would be less.
Answer: growth factor = 1.022
Step-by-step explanation:
The expression 11(1.022)^t is an exponential expression where the rate of increase is 2.2%
What 1.022 represent in this equation is the growth factor of per capita gross domestic product (GDP) of the US
In this question, 10 people average running distance is 2.7miles. Then the sum of their running distance would be: 10 people * 2.7 miles/people= 27 miles.
The group then divided into 2, 4 people with 3 miles average and 6 people with unknown average. Since the people are same like the 10 people group, their total running distance would be the same. The calculation would be:
total distance = group1 * average1 + group2 * average2
27 miles= 4 people * 3 miles/people + 6 people* average2
6 people * average2 = 27 miles -12 miles = 15 miles
average2= 15 miles/6people= 2.5 miles/people
For a set of data: x = (0,1,2,3,4,5,6) and y=(36, 28, 25, 24, 23, 21, 19), is it wise to use a linear regression to extrapolate
melisa1 [442]
Answer:
The problem with this solution is that a regression model is not recommended to extrapolate because we do not know if the linear relation that we calculated for a specific range of x values still holds outside this range.
Step-by-step explanation:
We have a linear regression model, with a range of the independent variable "x" that goes from 0 to 6.
The regression model finds a good fit (r=0.8582).
As it has a good fit, it is proposed to use this model to extrapolate and calculate the value of y for x=50.
It is not recommended to extrapolate a regression model unless we are really sure that the model is still valid within the range within we are extrapolating.
This means that if we have no proof that y has a linear relation in a range of x that includes x, the extrapolation has no validity and can lead to serious errors.
A linear regression model is only suitable for interpolation or extrapolating within the range we are sure that the relation between y and x is linear within a certain acceptable error.
