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Paladinen [302]
3 years ago
13

There are nine different marbles in a bag. Supposed you reach in and draw one at a time. And do this three times. How many ways

can you draw the three marbles if you do not replace the marble each time
Mathematics
1 answer:
Mumz [18]3 years ago
6 0

Answer: 504 ways

Step-by-step explanation:

Given

There are nine different marbles in a bag

For the first time, there are 9 possible ways to draw a marble

After removal of first marble, there are 8 possible ways to draw a marble

for the third time, there are 7 possible ways

so, total ways to draw three marble are

\Rightarrow 9\times 8\times 7=504\ \text{ways}

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Step-by-step explanation:

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Estimate the side length of a square that has a 9cm long diagonal
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S^2 + s^2 = 9^2 => 2s^2 = 81 => s^2 = 81/2 => side length s = 9(sqrt(2))/2 = 6.37 cm.
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Answer:

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Solve for x,,,3x-1/4=-5/11
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4 0
4 years ago
Answer questions 13. And 14. Please! Show minimal work...
ale4655 [162]

13. The equation of the given function is y=4sin[\frac{(t-\frac{4}{3})}{2}]-2

14. The equation of the cotangent function is y=cot[2(t-\frac{1}{3})]+2

Step-by-step explanation:

Let us revise the transformation of the trigonometric function:

y = a f[b(x + c)] + d, where

  • Amplitude is a
  • f represents the trigonometry function
  • Period is 2π/b
  • Phase shift is c (positive is to the left)
  • Vertical shift is d

13.

∵ y = a sin( \frac{2\pi t}{T} ), where a is the amplitude

   and T is the wave in seconds

∵ The amplitude is 4

∴ a = 4

∵ The period is 4π

∴ T = 4π

From the rules above

∵ The period is 2π/b

∴ T = \frac{2\pi }{B}

∴ 4π = \frac{2\pi }{b}

- By using cross multiplication

∴ 4π(b) = 2π

- Divide both sides by 4π

∴ b = \frac{1}{2}

∵ The phase shift is -\frac{4}{3}\pi

∵ c is the phase shift

∴ c = -\frac{4}{3}\pi

∵ The vertical shift is -2

∵ d is the vertical shift

∴ d = -2

Now substitutes the values of a, b, c and d in the form of the equation below

∵ y = a sin[b(t + c)] + d

∴ y=4sin[\frac{1}{2}(t-\frac{4}{3})]-2

You can write it as y=4sin[\frac{(t-\frac{4}{3})}{2}]-2

The equation of the given function is y=4sin[\frac{(t-\frac{4}{3})}{2}]-2

14.

y = cot[b(t + c)] + d

∵ The period = π

∵ The period is 2π/b

- Equate π by 2π/b to find b

∴ π = \frac{2\pi }{b}

- By using cross multiplication

∴ π(b) = 2π

- Divide both sides by π

∴ b = 2

∵ The phase shift is -\frac{1}{3}\pi

∵ c is the phase shift

∴ c = -\frac{1}{3}\pi

∵ The vertical shift is 2

∵ d is the vertical shift

∴ d = 2

Now substitutes the values of b, c and d in the form of the equation below

∵ y = cot[b(t + c)] + d

∴ y=cot[2(t-\frac{1}{3})]+2

The equation of the cotangent function is y=cot[2(t-\frac{1}{3})]+2

Learn more:

You can learn more about trigonometry function in brainly.com/question/3568205

#LearnwithBrainly

5 0
3 years ago
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