11. Any company plans to connect a total of 20 pF of capacitance to a 7200-V
1 answer:
You might be interested in
Part A
Purchase Value of Anna's Car in 2015 = $36,000
Depreciated Value of Anna's Car in 2017 = $25,000
We know that the value of the car is depreciating linearly.
From the above data, we can see that the value of the car has reduced by $36,000-$25,000 = $11,000 over 2 years.
⇒ Per year the value has been depreciated by
⇒ Per year the value has been depreciated by $5,500
So if the value of a car in 2015 is V₀ then value of the car after t years can be determined by the below function:
V (t) = V₀ - (t*5,500), where t indicates the years passed since 2015
OR V (t) = 36,000 - (t*5,500), where t indicates the years passed since 2015
Part B
Suggested value of Anna's car for trade in option in 2018 = $15,000
In 2018, 3 years would have been passed since 2015. Using the value of t = 3 in the function determined in Part A:
V(3) = V₀ - (3*5,500)
⇒ V(3) = 36,000 - (16,500)
⇒ V(3) = $19,000
From the above calculations, we can see that the value of the car in 2018 should be $19,000, however the suggested value for trade in is indicated as $15,000 which is lower than what the value of the car should be. Hence, basis linear depreciation, the $15,000 value is not fair for the car in 2018.
The value of the P(A∩B) is equal to 0.04.
We have given that,
P(A)= 0.4 and P(B) = 0.85.
We have to determine the P(A and B)
<h3>What is the formula for Independent Events?</h3>For Independent Events,
P(A) × P(B) = P(A∩B)
so we have, P(A∩B) = 0.4×0.1
P(A∩B) = 0.04
P(A') = 1 - 0.4 = 0.6
This information can be represented on a Venn diagram as shown below
P(A'∪B) means the union of everything that is not A with everything that is B
P(A'∪B) = 0.06 + 0.54 + 0.04
P(A'∪B) = 0.64
To learn more about the events visit:
#SPJ1
