All categories

2 years ago

Dan buys a car for £2200.

Mathematics

1 answer:

KatRina [158]2 years ago

8 0

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;

$FV = PV \cdot \left ( 1 + \dfrac{r}{100} \right )^n$

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;

$FV = 2200 \times \left ( 1 + \dfrac{-0.5}{100} \right )^6 \approx 2,134.82$

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))

You might be interested in

Which is more economical: purchasing the economy size of a detergent at 7 kilograms for $7.15 or purchasing the regular size a

Kruka [31]

Regular size saves $0.0003 per gram I think
someone pls fact check me

8 0

2 years ago

A construction company is considering submitting bids for contracts of three different projects. The company estimates that it h

julsineya [31]

Answer:

a. $P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

b. $E(x) = 0.3$

c. $S(x)=0.5196$

d. $E=5,000$

Step-by-step explanation:

The probability that the company won x bids follows a binomial distribution because we have n identical and independent experiments with a probability p of success and (1-p) of fail.

So, the PMF of X is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\$

Where p is 0.1 and it is the chance of winning. Additionally, n is 3 and it is the number of bids. So the PMF of X is:

$P(x)=\frac{3!}{x!(3-x)!}*0.1^{x}*(1-0.1)^{n-x}\\$

For binomial distribution:

$E(x)=np\\S(x)=\sqrt{np(1-p)}$

Therefore, the company can expect to win 0.3 bids and it is calculated as:

$E(x) = np = 3*0.1 = 0.3$

Additionally, the standard deviation of the number of bids won is:

$S(x)=\sqrt{np(1-p)}=\sqrt{3(0.1)(1-0.1)}=0.5196$

Finally, the probability to won 1, 2 or 3 bids is equal to:

$P(1)=\frac{3!}{1!(3-1)!}*0.1^{1}*(1-0.1)^{3-1}=0.243\\P(2)=\frac{3!}{2!(3-2)!}*0.1^{2}*(1-0.1)^{3-2}=0.027\\P(3)=\frac{3!}{3!(3-3)!}*0.1^{3}*(1-0.1)^{3-3}=0.001$

So, the expected profit for the company is equal to:

$E=-10,000+50,000(0.243)+100,000(0.027)+150,000(0.001)\\E=5,000$

Because there is a probability of 0.243 to win one bid and it will produce 50,000 of income, there is a probability of 0.027 to win 2 bids and it will produce 100,000 of income and there is a probability of 0.001 to win 3 bids and it will produce 150,000 of income.

5 0

2 years ago

Lin counts 5 bacteria under a microscope. She counts them again each day for four days, and finds

svetlana [45]

Answer:

Yes, but it is not linear. It is exponential : $f(x) = 5\cdot 2^{x-1}$