2.A 1998 Pontiac Grand-Am depreciates in value by 18% on average each year. If the car originally sold for $19995 in 1998, how m
1 answer:
Answer:
Amount of car in 2012 = $1,242.55 (Approx)
Step-by-step explanation:
Given:
Rate of depreciation(d) = 18% = 0.18
Amount of car in 1998 = $19,995
Find:
Amount of car in 2012
Computation:
Number of year(n) = 14 year
Amount of car in 2012 = 19,995[1-0.18]¹⁴
Amount of car in 2012 = 19,995[0.82]¹⁴
Amount of car in 2012 = 19,995[0.0621432458]
Amount of car in 2012 = 1,242.5542
Amount of car in 2012 = $1,242.55 (Approx)
You might be interested in
Answer:
Z-score 1.96
Margin of error d= 0.04323
Step-by-step explanation:
Hello!
The study variable in this case is
X: Number of people that oppose the use of photo-cop for issuing traffic tickets in a sample of 500.
n= 500
The parameter of interest is the proportion of people that opposes it.
The estimated proportion is p'= 300/500= 0.6
To estimate the population proportion per Confidence interval you have to approximate the distribution of the sample proportion p' to normal:
p'≈N(p;p(1-p)1/n)
The mean of this distribution is p and the variance is p*(1-p)*1/n
To construct the Confidence interval, since the value of the population proportion is unknown, an estimated variance is used:
p'(1-p')*1/n ⇒ then the estimated standard error is √(p'(1-p')*1/n)
The formula for the confidence interval is:
p' ± *
Where " *
" represents the margin of error of the interval.
Now for a confidence level of 0.95 the value of Z is
The estimated standard error is already calculated:
The margin of error (d) is then:
d= *
= 1.965 * 0.022
d= 0.04323
I hope it helps!