Answer:
10.24 years
Step-by-step explanation:
The computation of the time period is as follows:
As we know that
Amount = Principal × (1 + rate of interest)^time period
£8000 = £4000 × (1 + 0.07)^time period
£8,000 ÷ £4,000 = 1.07^time period
2 = 1.07^time period
Now apply the log to the both sides
log 2 ÷ log 1.07 = time period
= 10.24 years
Answer:
because 6 is larger then 4
Step-by-step explanation:
up
The question involves the concept & equations associated with projectile motion.
Given:
y₁ = 1130 ft
v₁ = +46 ft/s (note positive sign indicates upwards direction)
t = 6.0 s
g = acceleration due to gravity (assumed constant for simplicity) = -32.2 ft/s²
Of the possible equations of motion, the one we'll find useful is:
y₂ = y₁ + v₁t + 1/2gt²
We can just plug and chug to define the equation of motion:
<u><em>y = (1130 ft) + (46 ft/s)t + 1/2(-32.2 ft/s²)t²</em></u>
<em>(note: if you were to calculate y using t = 6.0 s, you'd find that y = 826.4 ft, instead of 830 ft exactly because of some rounding of g and/or the initial velocity)</em>
For a smoothing constant of 0.2
Time period – 1 2 3 4 5 6 7 8 9 10
Actual value – 46 55 39 42 63 54 55 61 52
Forecast – 58 55.6 55.48 52.18 50.15 52.72 52.97 53.38 54.90
Forecast error - -12 -.6 -16.48 – 10.12 12.85 1.28 2.03 7.62 -2.9
The mean square error is 84.12
The mean forecast for period 11 is 54.38
For a smoothing constant of 0.8
Time period – 1 2 3 4 5 6 7 8 9 10
Actual value – 46 55 39 42 63 54 55 61 52
Forecast – 58 48.40 53.68 41.94 41.99 58.80 54.96 54.99 59.80
Forecast error - -12 6.60 -14.68 0.06 21.01 -4.80 0.04 6.01 -7.80The mean square error is 107.17
The mean forecast for period 11 is 53.56
Based on the MSE, smoothing constant of .2 offers a better model since the mean forecast is much better compared to the 53.56 of the smoothing constant of 0.8.
