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dangina [55]
3 years ago
10

The value of a car depreciates by 18% per year. Work out the current value of a car bought 4 years ago for £20000

Mathematics
2 answers:
dem82 [27]3 years ago
5 0
Given:
depreciation rate: 18%
Value of the car : 20,000
Age of the car : 4 years.

The depreciation is based on the current value of the car. Therefore, the amount of depreciation varies.

<span> <span> </span><span><span> yr   Beginning Value   Dep. Rate    Depreciation      Ending Value
</span> <span> 1 <span>   20,000.00                 </span>18% <span>             3,600.00 </span> <span>         16,400.00
</span> </span> <span> 2 <span>   16,400.00                 </span>18% <span>             2,952.00       </span> <span>   13,448.00
</span> </span> <span> 3 <span>   13,448.00                 </span>18% <span>             2,420.64 </span> <span>         11,027.36
</span> </span> <span> 4 <span>   11,027.36                 </span>18% <span>             1,984.92 </span> <span>           9,042.44
</span> </span></span></span>
The current value of a car bought 4 years ago is 9,042.44

Beginning value : purchased amount on 1st year. then, ending balance of the previous year from 2nd year onwards.
Depreciation : Beginning value * depreciation rate
Ending value : Beginning value - depreciation

ASHA 777 [7]3 years ago
5 0
20000 = 100% 18 x 4 = 72% 20000 / 100 = 200 200 = 1% 200 x 72 = 14,400 20000 - 14,400 = 5600 The value of it now should be £5600
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A researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic
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Answer:

90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

Step-by-step explanation:

We are given that a researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic meter with a standard deviation of 0.093.

Firstly, the pivotal quantity for 90% confidence interval for the population mean iron concentration is given by;

         P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean iron concentration = 0.802 cc/cubic meter

             s = sample standard deviation = 0.093

             n = number of water samples = 27

             \mu = population mean

<em>Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 90% confidence interval for the population mean, \mu is ;

P(-1.706 < t_2_6 < 1.706) = 0.90  {As the critical value of t at 26 degree of

                                                 freedom are -1.706 & 1.706 with P = 5%}

P(-1.706 < \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } < 1.706) = 0.90

P( -1.706 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 1.706 \times {\frac{s}{\sqrt{n} } } ) = 0.90

P( \bar X -1.706 \times {\frac{s}{\sqrt{n} } < \mu < \bar X +1.706 \times {\frac{s}{\sqrt{n} } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X -1.706 \times {\frac{s}{\sqrt{n} } , \bar X +1.706 \times {\frac{s}{\sqrt{n} } ]

                                                 = [ 0.802 -1.706 \times {\frac{0.093}{\sqrt{27} } , 0.802 +1.706 \times {\frac{0.093}{\sqrt{27} } ]

                                                 = [0.771 , 0.832]

Therefore, 90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

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vazorg [7]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Hope this helps.

8 0
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