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alukav5142 [94]
3 years ago
14

A single die is rolled twice. find the probability of rolling an odd number the first time and a number greater than 3 the secon

d time. find the probability of rolling an odd number the first time and a number greater than 3 the second time.
Mathematics
1 answer:
Lisa [10]3 years ago
8 0

|\Omega|=6\cdot6=36\\ |A|=3\cdot3=9\\\\ P(A)=\dfrac{9}{36}=\dfrac{1}{4}=25\%

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(4x1000) + (2x100) + (3x1/10)
disa [49]
(4*1000) + (2*100) + (3*1/10) = <span>4200.3
</span>(4x1000) + (2x100) + (3x1/10) = <span>4200.3
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Both are equal.
5 0
3 years ago
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How do i solve 426 to the nearest ten
Dmitry [639]
Youll need to provide more information
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Need help on these !! 13 and 14! Please!!
katen-ka-za [31]

Question # 13

Answer:

The required equation for the given function is <em>y = 4sin(x/2+2π/3) -2 , as shown attached graph diagram.</em>

<em>Step-by-step explanation: </em>

As the general sine function is given by

y=asin(bx+c)+d.......[A]

  • amplitude = a
  • period = 2π ÷ b
  • Phase shift = -c ÷ b
  • Vertical shift = d

As in the question,  

  • amplitude = a = 4
  • period = 4π
  • phase shift = -4π/3
  • Vertical shift = d = -2

As  

period = 2π ÷ b  

b = 2π/period

b = 2π/4π ∵ period = 4π

b = 1/2  

Also

Phase shift = -c/b

-4π/3 = -c/b ∵ phase shift = -4π/3

4π/3 = c/b  

c = b × 4π/3  

c = 1/2 × 4π/3  

c = 4π/6  

c = 2π/3

So, putting Amplitude ⇒ a = 4, Vertical shift ⇒ d = -2, b = 1/2 ,  

and c = 2π/3 in Equation [A] would bring us the required equation for the given function.

y=asin(bx+c)+d

y = 4sin(x/2+2π/3)+(-2)

y = 4sin(x/2+2π/3) -2            

<em>Note: The graph is also shown in attached diagram.</em>

                                             Question # 14

<em>Answer:</em>

The required equation for the given function is y = cot(x+π/3)+2, as shown in attached graph diagram.

<em>Step-by-step explanation: </em>

As the general cotangent function is given by

y=acot(bx+c)+d.......[A]

  • amplitude = a
  • period = π ÷ b
  • Phase shift = -c ÷ b
  • Vertical shift = d

As in the question,  

  • period = π
  • phase shift = -π/3
  • Vertical shift = d = 2

As  

period = π ÷ b  

b = π/period

b = π/π ∵ period = 4π

b = 1  

Also

Phase shift = -c/b

-π/3 = -c/b ∵ phase shift = -π/3

π/3 = c/b  

c = b × π/3  

c = 1 × π/3  

c = π/3

So, putting vertical shift ⇒ d = 2, b = 1 and   c = π/3 in Equation [A] would bring us the required equation for the given function.

y=acot(bx+c)+d

y = cot(x+π/3)+2

<em>Note: The graph is also shown in attached diagram.</em>

Keywords: amplitude, period , phase shift , vertical shift

Learn more about trigonometric functions of equations from brainly.com/question/2643311

#learnwithBrainly

7 0
3 years ago
If you are good at algebra 2 please help me with this question.
AysviL [449]
Where’s the question?
3 0
3 years ago
3x2=7 5x4=23 7x6=47 9x8=79 10x9=?
Mashcka [7]
(a\times b) = (a\times b) +(b -1) = c \\  \\ 3 \times 2 = (3\times 2) +(2-1) = 6 +1=7 \\  \\ 5 \times 4 = (5\times 4) +(4-1) = 20 +3=23 \\  \\ 7 \times 6 = (7\times 6) +(6-1) = 42 +5=47 \\  \\ 9 \times 8= (9\times 8) +(8-1) = 72 +7=79 \\  \\  \\  \\ 10 \times 9 = (10\times 9) +(9-1) = 90 +8=\boxed {\bf{98}}\\\sf {10\times 9 = \boxed{\bf {98}}}
3 0
2 years ago
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