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1 year ago

Dan buys a car for £2200.

Mathematics

1 answer:

KatRina [158]1 year 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))

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<u>Statement </u> <u>Reason </u>

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1 year ago

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Tju [1.3M]

Answer:

a) $K(t)=6\frac{tie}{min}(t)$

b) $J(t)=11\frac{tie}{min}(t)$

c) $T_{f}(t)=K(t+10)+J(t)$ where $T_{f}(t)$ = number ties finished per unit time

d) t≅8.24min

e) K=109tie

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f) K≅16.6min

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Step-by-step explanation:

Data

$Kara_{(K)}=6\frac{tie}{min}\\Julie(J)=11\frac{tie}{min}$

Kara begins ten minutes before julie, so when julie joins kara she has 60 ties

t=10 $k=6\frac{tie}{min}*10min=60tie$

a) $K(t)=6\frac{tie}{min}(t)$

b) $J(t)=11\frac{tie}{min}(t)$

c) $T_{f}(t)=K(t+10)+J(t)$ where $T_{f}(t)$ = number ties finished per unit time

d) if kara already has 60 ties and makes 6 per minute and Julie makes 11 per minute then per minute they makes 17 ties, so one minute per 17 ties for 200 ties ¿how long?. $t=\frac{(200-60)tie*min}{17tie}$ ≅8.24min

e) K=60tie+6tie/min*8.24min=109tie

J=11tie/min*8.24min=91tie

f) K=100tie*min/6tie≅16.6min

J=100tie*min/11tie≅9.09min

Note: the amount of ties were rounded to the nearest decimal

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