Ken's distance is: d(t)=1.7t
Ken's shadow is:
(14-6)/d(t)=14/s
8/d(t)=14/s
s=14d(t)/8
s=1.75d(t) and using the value for d(t) we have:
s=1.75(1.7t)
s=2.975t and if you want it in terms of d, d=1.7t, t=d/1.7 so
s=2.975(d/1.7)
s=1.75d
First we will find the interest on:
P = $235 principal
t = 2 years
r = 0.1415 annual rate
A = future value
I = A - P the interest
A = P(1 + r)^t
A = 235(1 + 0.1415)^2
A = $306.21
I = A - P
I = $306.21 - $235
I = $71.21
the interest was $71.21.
Next lets find the lifetime cost value:
Lifetime cost value = 306.21 + 5*1.56*52 + 5*0.78*52 = $914.61 (considering that 1 year = 52 weeks)
Now lets find the percentage what percentage the interest is of the lifetime cost:
(71.21/914.61)*100 = 7.79%
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Answer:
As the sample size increases, the variability decreases.
Step-by-step explanation:
Variability is the measure of actual entries from mean. The less the deviations the less would be the variance.
For a sample of size n, we have by central limit theorem the mean of sample follows a normal distribution for random samples of large size.
X bar will have std deviation as 
where s is the square root of variance of sample
Thus we find the variability denoted by std deviation is inversely proportion of square root of sample size.
Hence as sample size increases, std error decreases.
As the sample size increases, the variability decreases.
Answer:
Step-by-step explanation:
(-3 ab²)³=(-3)³(a)³(b²)³=-27 a³b^(2×3)=-27 a³b^6
Answer:
102 marbles
Step-by-step explanation:
100% of 85 is 85.
20% of 85 = 17
85 + 17 = 102
Hope that helps!
