Anne-Marie Cole runs the sales division for a local auto insurance firm. One of her key duties is to calculate her company's mar
2 answers:
Answer:
The current share of Anne-Marie's in market is 10%.
Explanation:
Achieve total sales = $ 3 million
Entire industry sales = $ 30 million
Lets find the share of Anne-Marie's in the current market
The share of Anne-Marie's in market is S
Now by putting the values in the above formula
S= 0.1
Percentage S= 0.1 x 100
S =10 %
So the current share of Anne-Marie's in market is 10%.
You might be interested in
Solution :
The percentage of the students who have a chance of repeating their current year = 3%
The drop out students for the first year and the sophomores = 6%
Drop out rate of first year and the seniors = 4%
Now for the state space :
S = { first year(1), sophomores(2), juniors(3), seniors(4), graduates(G), Dropouts(D) }
Therefore
the first year students are indicated as '1'
Sophomores are indicated as '2'
Juniors are indicated as '3'
Seniors are indicated as '4
Graduates are indicated as 'G'
Dropouts are indicated as 'D'
The transition diagram is attached below.
The probability of the students who have the chance of repeating their current year = 3/100 = 0.03
Probability of first year dropouts and sophomores = 6/100 = 0.06
Probability of dropout rate of juniors and seniors = 4/100 = 0.04
Therefore, the probability matrix can be made as :
1 2 3 4 G D
Here, G represents 'graduates' and D represents 'Dropouts.'
Answer:
a) I=0 b) 4.17V c) 0.354 A d) 14.5s
Explanation:
a) consider circuit in the attachment
i(t)= E/R (1- e^(-t/RL))
i(0)= 12.5/3×(1-e^(0/RL))
i(0)=0
b) at t⇒∞
i(∞)= 12.5/3× (1- e^(-∞/RL))
= 4.17V
c) 1/RL= 1/(6.95×3)= 0.0479616
i(1.85) = 12.5/3 × (1- e^(-1.85×0.0479616)
= 0.354A
d) I/2= I (1- e^(-t/RL))
t= - RL ln0.5
t= - 3×6.95 × (-0.693)
t= 14.5 s
