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jonny [76]
4 years ago
7

If a company issues 2,500,000 shares with voting rights how many shares must an investor by to be assured control of the company

Mathematics
1 answer:
Delvig [45]4 years ago
3 0
B is the answer i beliebe
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Translate to an algebraic expression:<br><br> 12 less than the square of a number
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Answer:

12-x²

Step-by-step explanation:

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3 years ago
Jill is standing at the base of her apartment building. She measures the angle of elevation to the top of a nearby tower to be 4
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Answer:

1) From the measure of 40°, you can write:

tan(40°) = 100/x, where x is the base from the building to the tower

⇒x=100/tan(40°) = 119,18 m

2) From the measure of 30°, you can write

tan(30°) = y / 119,18, where y is the height from the roof of Jill's building to the top of the tower.

Then, y = tan(30°) * 119,18 = 68,81 m

3) The height of Jill's building is 100 - 68,81 = 31,19 m

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4 years ago
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Answer:

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Step-by-step explanation:

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7 0
3 years ago
Read 2 more answers
In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the
earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0
3 years ago
I need help with this one please help meee
Ket [755]
Sheeeeeeeeeeesh I feel bad for you
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3 years ago
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