Answer:
The worth of the car after 6 years is £2,134.82
Step-by-step explanation:
The amount at which Dan buys the car, PV = £2200
The rate at which the car depreciates, r = -0.5%
The car's worth, 'FV', in 6 years is given as follows;

Where;
r = The depreciation rate (negative) = -0.5%
FV = The future value of the asset
PV = The present value pf the asset = £2200
n = The number of years (depreciating) = 6
By plugging in the values, we get;

The amount the car will be worth which is its future value, FV after 6 years is FV ≈ £2,134.82 (after rounding to the nearest penny (hundredth))
Answer:
The 99% confidence interval for the average length of time all car owners plan to keep their cars is between 3.85 years and 10.55 years.
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so 
Now, find the margin of error M as such

In which
is the standard deviation of the population and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 7.2 - 3.35 = 3.85 years
The upper end of the interval is the sample mean added to M. So it is 7.2 + 3.35 = 10.55 years
The 99% confidence interval for the average length of time all car owners plan to keep their cars is between 3.85 years and 10.55 years.
Answer: The mercury is 38 mi far from sun.
Step-by-step explanation:
Let x denotes the distance ( in mi )between the Sun and Mercury .
Distance from Venus to Sun = x+30.9
Distance from Earth = x+30.9+ 23.6 =x+30.9+ 23.6 =x+54.5
When the total of the distances for these plants from the sun is 199.4 mi.
According to the given statements , we have



[Divide both sides by 3]
Hence, the mercury is 38 mi far from sun.