Answer:
and 
Step-by-step explanation:
Given
See attachment for complete question
Required
Determine the equilibrium solutions
We have:


To solve this, we first equate
and
to 0.
So, we have:


Factor out R in 

Split
or 
or 
Factor out W in 

Split
or 
Solve for R


Make R the subject


When
, we have:




Collect like terms

Solve for W




When
, we have:



Collect like terms

Solve for R


So, we have:

When
, we have:





So, we have:

Hence, the points of equilibrium are:
and 
Answer:
The minimum level for which the battery pack will be classified as highly sought-after class is 2.42 hours
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 is the minimum level for which the battery pack will be classified as highly sought-after class
At least the 100-10 = 90th percentile, which is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.




The minimum level for which the battery pack will be classified as highly sought-after class is 2.42 hours
Answer:
eqn of line
3x-4y=-16
Step-by-step explanation:
Using compound interest, the rates per compounding period are given as follows:
a) 0.1273 = 12.73%.
b) 0.0833 = 8.33%
c) 0.0617 = 6.17%
<h3>What is compound interest?</h3>
The amount of money earned, in compound interest, after t years, is given by:

In which:
- A(t) is the amount of money after t years.
- P is the principal(the initial sum of money).
- r is the interest rate(as a decimal value).
- n is the number of times that interest is compounded per year.
The <u>interest rate per compounding period</u> is given as follows:

For item a, the parameters are:
r = 0.12, n = 52.
Hence:

For item b, the parameters are:
r = 0.08, n = 104.
Hence:

For item c, the parameters are:
r = 0.06, n = 12.
Hence:

More can be learned about compound interest at brainly.com/question/25781328
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Answer: Reshma is making a necklace using green beads and purple beads in a ratio represented on the following double number line. Fill in the missing values on the diagram.
If Reshma uses 20 green beads, how many purple beads will she use?