Answer:
The bulbs should be replaced each 1060.5 days.
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:

How often should the bulbs be replaced so that no more than 1% burn out between replacement periods?
This is the first percentile, that is, the value of X when Z has a pvalue of 0.01. So X when Z = -2.325.




The bulbs should be replaced each 1060.5 days.
Answer:
y-b/m = x
Step-by-step explanation:
y = mx + b
~Subtract b to both sides
y - b = mx
~Divide m to both sides
y-b/m = x
Best of Luck!
<span>First year is 30,000.
</span><span>Earn 5% raise every year.
</span>Growth factor is 1.05 this sounds a lot like a geometric ratio.
An = A1 × r^(n-1)Sn = A1 × (1 - r^n) / (1-r)<span>n = 40
A1 = 30,000
A40 = $30,000 * 1.05^39 = $201,142.53
S40 = A1 * (1 - 1.05^40) / (1-1.05) which becomes:
S40 = 30,000 * -6.039988712 / -.05 which becomes:
</span>S40 = -181,199.6614 / -.05 which becomes:
S40 = $3,623,993.227
<span>The individual yearly calculations are shown below: </span>
F(x) = 5x - 2
f(-3/5) = 5(-3/5) - 2
f(-3/5) = -15/5 - 2
f(-3/5) = -3 - 2
f(-3/5) = -5
C
I think this is the right answer