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Scorpion4ik [409]

3 years ago

13

a stock that was selling for $48 a share split 2-for-1. Before the split, the company had 3.4 million shares of stock outstandin

g. What is the post-split number of shares outstanding?

2 answers:

hjlf3 years ago

6 0

I’m not sure

I’m sorry

nalin [4]3 years ago

4 0

Answer:

666

Step-by-step explanation:

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3 years ago

What is the answer

aliya0001 [1]

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Step-by-step explanation:

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3 0

3 years ago

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Lori was paid a 15 percent commission for the home audio system that she sold. The electronics store had marked up the price of

svlad2 [7]

Answer:

$63.36

Step-by-step explanation:

The home audio system costs $264 wholesale.

The system's price had been increased by 60 percent at the electronics retailer.

Consequently, the increased cost is 60/100x264=$158.40.

The audio system's total cost is now $422.40 (264 + 158.40).

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2 years ago

Which operation should be performed first according to the order of operations?

iragen [17]

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3 years ago

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The probability that a single radar station will detect an enemy plane is 0.65.

taurus [48]

Answer:

a) We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

b) If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

Step-by-step explanation:

For each station, there are two two possible outcomes. Either they detected the enemy plane, or they do not. This means that we can solve this problem using concepts of the binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

The probability that a single radar station will detect an enemy plane is 0.65. This means that $n = 0.65$ .

(a) How many such stations are required to be 98% certain that an enemy plane flying over will be detected by at least one station?

This is the value of n for which $P(X = 0) \leq 0.02$ .

n = 1.

$P(X = 0) = C_{1,0}.(0.65)^{0}.(0.35)^{1} = 0.35$

n = 2

$P(X = 0) = C_{2,0}.(0.65)^{0}.(0.35)^{2} = 0.1225$

n = 3

$P(X = 0) = C_{3,0}.(0.65)^{0}.(0.35)^{3} = 0.0429$

n = 4

$P(X = 0) = C_{4,0}.(0.65)^{0}.(0.35)^{4} = 0.015$

We need 4 stations to be 98% certain that an enemy plane flying over will be detected by at least one station.

(b) If seven stations are in use, what is the expected number of stations that will detect an enemy plane?

The expected number of sucesses of a binomial variable is given by:

$E(x) = np$

So when $n = 7$

$E(x) = 7*(0.65) = 4.55$

If seven stations are in use, the expected number of stations that will detect an enemy plane is 4.55.

6 0

3 years ago