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harina [27]
2 years ago
15

Suppose the probability of being picked first in your class is 0.05. What are the the odds of being picked first? Express your a

nswer in the form a:b
Mathematics
1 answer:
den301095 [7]2 years ago
5 0

Suppose the probability of being picked first in your class is 0.05. The odds of being picked first expressed in form of a: b is 1 : 19.

A probability is a number that represents the possibility or chance that a specific event will occur.

{\mathbf{\text{The probability of an event} = \dfrac{\text{number of desired outcome}}{\text{total number of possible outcome}}}

From the given information;

  • Let assume the total number of people in the class is whole = 1

∴

The odds of being picked first can be computed as:

\mathbf{=\dfrac{0.05}{1-0.05}}

\mathbf{=\dfrac{0.05}{0.95}}

\mathbf{\dfrac{1}{19}}

= 1 : 19

Therefore, the odds of being picked first in terms of a: b is 1 : 19

Learn more about probability here:

brainly.com/question/11234923?referrer=searchResults

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3 years ago
Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps
Firdavs [7]
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
   x²(-3 - x²) = -4
   -3x² - x⁴ = -4
   x⁴ + 3x² - 4 = 0
   (x² + 4)(x² - 1) = 0
   (x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
   y = -3 - (±1)²
   y = -3 - 1
   y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

   -3 + ix²y 
   = -3 + i(±1)²(-4)
   = -3 - 4i

   x² + y + 4i
   = (±1)² - 4 + 4i
   = 1 - 4 + 4i
   = -3 + 4i

They result in conjugates
4 0
3 years ago
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