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ANEK [815]
1 year ago
8

. A manager has just received the expense checks for six of her employees. She randomly distributes the checks to the six employ

ees. What is the probability that exactly five of them will receive the correct checks (checks with the correct names)?
Mathematics
1 answer:
Tcecarenko [31]1 year ago
3 0

Answer:

13,8%

Step-by-step explanation:

There are six employees and six cheks, so there are 36 (6x6) possible combinations so if we need to measure the probability that five of them receive the exact check  is only one for each one of them over the 36 possibilities, so 1/36 for one plus 1/36 the second and so on.  5/36 = 13,8%.

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The ABC sequence of points is clockwise in both figures, so there will be an even number of reflections or a rotation.

Rotation 90° clockwise about the point (-3, -3) would make the required transformation, but that is not an option. An equivalent is ...
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This isosceles triangle has two sides of equal length, a, that are longer than the length of the base, b. The perimeter of the t
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Isosollese triangle has 2 equal sides that are longer than legnth of base
perimiter=15.7
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8 0
1 year ago
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The closing price (in dollars) per share of stock of tempco electronics on the tth day it was traded is approximated by p(t) = 2
Vaselesa [24]

solution:

The closing price (in dollars) per share of stock of Tempco Electronics on the tth day it was  

traded is approximated by  

P(t) = 20 + 12 sin πt/30 − 6 sin πt/15 + 4 sin πt/10 − 3 sin 2πt/15 (0 ≤ t ≤ 24)  

where t = 0 corresponds to the time the stock was first listed on a major stock exchange.  

What was the rate of change of the stock's price at the close of the 15th day of trading?  

P'(t) = 12(cos πt/30) - 6 (cos πt/15) + 4(cos πt/10) - 3(cos 2πt/15)  

t = 15  

P'(t) = 12(cos 15π/30) - 6 (cos 15π/15) + 4(cos 15π/10) - 3(cos 30π/15)  

P'(t) = 12(cos π/2) - 6 (cos π) + 4(cos 3π/2) - 3(cos 2π)  

P'(t) = 12(0) - 6(-1) + 4(0) - 3(1) = 6 - 3  

P'(t) = $3 per day RATE OF CHANGE  

the closing price on that day

P(t) = 20 + 12 sin πt/30 − 6 sin πt/15 + 4 sin πt/10 − 3 sin 2πt/15  

t = 15  

P(t) = 20 + 12 sin 15π/30 − 6 sin 15π/15 + 4 sin 15π/10 − 3 sin 30πt/15  

P(t) = 20 + 12 sin π/2 − 6 sin π + 4 sin 3π/2 − 3 sin 2π  

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P(t) = $28 per share close price



4 0
1 year ago
A. [4 marks]
storchak [24]

The value of d is -2

Step-by-step explanation:

The nth term of the arithmetic sequence is u_{n}=u_{1}+(n-1)d , where

  • u_{1} is the first term
  • d is the common difference between each 2 consecutive terms

The nth term of the geometric sequence is u_{n}=u_{1}r^{n-1} , where

  • u_{1} is the first term
  • r is the common ratio between each two consecutive terms r=\frac{u_{2}}{u_{1}}=\frac{u_{3}}{u_{2}}

∵ u_{1} of an arithmetic sequence is 1

∵ The common difference is d, where d ≠ 0

∵ u_{2} , u_{3} , u_{6} are the first 3 terms of a geometric sequence

∵ u_{2} = 1 + (2 - 1)d

∴ u_{2} = 1 + d

∵ u_{3} = 1 + (3 - 1)d

∴ u_{2} = 1 + 2d

∵ u_{6} = 1 + (6 - 1)d

∴ u_{2} = 1 + 5d

∴ The first 3 terms of the geometric sequence are (1 + d) , (1 + 2d) ,

   (1 + 5d)

∵ The common ratio in the geometric sequence is r=\frac{u_{2}}{u_{1}}=\frac{u_{3}}{u_{2}}

∴ r=\frac{1+2d}{1+d}=\frac{1+5d}{1+2d}

- Use cross multiplication with the equal fractions

∵ \frac{1+2d}{1+d}=\frac{1+5d}{1+2d}

∴ (1 + 2d)(1 + 2d) = (1 + d)(1 + 5d)

∴ 1 + 2d + 2d + 4d² = 1 + 5d + d + 5d²

- Add like terms in each side

∴ 1 + 4d + 4d² = 1 + 6d + 5d²

- Subtract 1 from both sides

∴ 4d + 4d² = 6d + 5d²

- Subtract 4d from both sides

∴ 4d² = 2d + 5d²

- Subtract 4d² from each side

∴ 0 = 2d + d²

- Take d as a common factor

∴ 0 = d(2 + d)

- Equate each factor by 0

∴ d = 0 but we will reject it because d ≠ 0

∴ 2 + d = 0

- Subtract 2 from both sides

∴ d = -2

The value of d is -2

Learn more:

You can learn more about the sequence in brainly.com/question/1522572

#LearnwithBrainly

7 0
2 years ago
Justify each steps for X/3-7=11
wel
\dfrac{x}{3}-7=11\ \ \ \ |add\ 7\ to\ both\ sides\\\\\dfrac{x}{3}=18\ \ \ \ \ |multiply\ both\ sides\ by\ 3\\\\\boxed{x=54}
6 0
2 years ago
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