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Ksju [112]
3 years ago
5

A carnival game wheel has 12 equal sections. One of the sections contains a star . To win a prize the player must land on the se

ction with the star on two
consecutive spins. What is the probability of a player winning?


Please give a detailed answer
Mathematics
1 answer:
krok68 [10]3 years ago
7 0
The probability of landing a star is 1/12 and when you spin it again because nothing has changed it is still 1/12 to find the probability you would multiply them. So 1/12 x 1/12 = 1/144
the probability of landing on a star two consecutive times is a 1/144 chance.
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Answer:

\huge\boxed{\sf a = -12 , b = 8}

Step-by-step explanation:

<u>Given polynomial is:</u>

f(x) = ax^3+ b^2 + 3x + 4

\rule[225]{225}{2}

First factor is (x-1)

Let x - 1 = 0  => x = 1

Putting this in the above polynomial

f(1) = a(1)^3 + b(1)^2 + 3(1) + 4

Given that it's remainder is 3

3 = a + b + 3 + 4

3 = a + b + 7

Subtracting 7 to both sides

-4 = a + b\\OR \\

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\rule[225]{225}{2}

Second Factor is 2x + 1

Let 2x + 1 = 0 => x = - 1 / 2

Putting in the above polynomial

f(-\frac{1}{2} ) = a (-\frac{1}{2} )^3 + b (-\frac{1}{2} )^2  + 3 (-\frac{1}{2} ) + 4

The remainder is 6

6 = -\frac{a}{8} + \frac{b}{4} - \frac{3}{2} + 4\\6 - 4 = -\frac{a}{8} + \frac{b}{4} - \frac{3}{2}\\2 = \frac{-a + 2b -12 }{8} \\-a + 2b - 12 = 16\\-a + 2b = 16 + 12\\

<h2>-a + 2b = 28  --------------------------------(2)</h2>

\rule[225]{225}{2}

Adding both equations

a + b -  a + 2b = -4 + 28

3b = 24

Dividing both sides by 3

b = 8

\rule[225]{225}{2}

Putting b = 8 in Equation (1)

a + 8 = -4

a = - 4 - 8

a = -12

\rule[225]{225}{2}

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soldi70 [24.7K]

Answer:

Yes

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