7.50
25% also equals 1/4 so divide 130.00 by 4 which equals 32.50
subtract 32.50 from 130.00 which equals the cost, 97.50
97.50 minus 90$ is 7.50, therefore that is the answer
Each colection day: D
Number of tops collected on that day: N
D1=1; N1=2
D2=3; N2=8
1) Linear model
N-N1=m(D-D1)
m=(N2-N1)/(D2-D1)
m=(8-2)/(3-1)
m=(6)/(2)
m=3
N-N1=m(D-D1)
N-2=3(D-1)
N-2=3D-3
N-2+2=3D-3+2
N=3D-1
when D=6:
N=3(6)-1
N=18-1
N=17
<span>What is the number of tops collected on the sixth day based on the linear model?
</span>The number of tops collected on the sixth day based on the linear model is 17.
2) Exponential model
N=a(b)^D
D=D1=1→N=N1=2→2=a(b)^1→2=ab→ab=2 (1)
D=D2=3→N=N2=8→8=a(b)^3→8=a(b)^(1+2)
8=a(b)^1(b)^2→8=ab(b)^2 (2)
Replacing (1) in (2)
(2) 8=2(b)^2
Solving for b:
8/2=2(b)^2/2
4=(b)^2
sqrt(4)=sqrt( b)^2 )
2=b
b=2
Replacing b=2 in (1)
(1) ab=2
a(2)=2
Solving for a:
a(2)/2=2/2
a=1
Then, the exponential model is N=1(2)^D
N=(2)^D
When D=6:
N=(2)^6
N=64
<span>What is the number of tops collected on the sixth day based on the exponential model?
</span><span>The number of tops collected on the sixth day based on the exponential model is 64</span>
Answer:
The company should promote a lifetime of 3589 hours so only 2% burnout before the claimed lifetime
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What lifetime should the company promote for these bulbs, whereby only 2% burnout before the claimed lifetime?
This is the value of X when Z has a pvalue of 0.02. So it is X when Z = -2.055.




The company should promote a lifetime of 3589 hours so only 2% burnout before the claimed lifetime
A = 3x
b = 2x+25
180 = a + b
180 = 3x + 2x + 25
180 = 5x + 25
155 = 5x
x = 31
a = 3*31 = 93
b = 2*31 + 25 = 87