Answer:
£14889.30
Step-by-step explanation:
The cost of the car = £15500.
For the first year,
depreciation = £15500 x 1%
= £155
Its worth after the first year = £15500 - £155
= £15345
For the 2nd year,
depreciation = £15345 x 1%
= £153.45
Its worth after the second year = £15345 - £153.45
= £15191.55
For the 3rd year,
depreciation = £15191.55 x 1%
= £151.9155
Its worth after the third year = £15191.55 - £151.9155
= £15039.6345
For the 4th year,
depreciation = £15039.6345 x 1%
= £150.3963
Its worth after the fourth year = £15039.6345 - £150.3963
= £14889.2382
Thus, the worth of the car in 4 years would be £14889.30
Answer:
first is 6/8 and second is 7/8
Step-by-step explanation:
You really just have to round the numbers. Its super simple what number would round to the number that they are saying they are?
By looking at the bottom length of both triangles, we can determine the scale from one to the other.
C-E = 30 units
D-F = 28 units
This can be translated as the fraction 15/14.
So if we look back at the possible answers, both B and C are likely candidates for being the answer.
But to get the answer, we multiply 24 X (15/14).
Which is 25.7, or 25 and 5/7. Your answer is C.
The answer to this probability question is .86 or 86%. The probability that a batch of 10,000 batteries will contain at least 175 defective batteries is 86 percent. I calculated the failure rate of 10,000 batteries (10k*.015) and yielded 150, meaning there will be 150 batteries that will be defective in a set of 10,000. Then i divided 150 by 175 to get the probability.
