Answer:
Malcolm is showing evidence of gambler's fallacy.
This is the tendency to think previous results can affect future performance of an event that is fundamentally random.
Step-by-step explanation:
Since each round of the roulette-style game is independent of each other. The probability that 8 will come up at any time remains the same, equal to the probability of each number from 1 to 10 coming up. That it has not come up in the last 15 minutes does not increase or decrease the probability that it would come up afterwards.
Solve algebrically 3x - 4y = -24 and x + 4y = 8 is x = -4 and y = 3
<u>Solution:</u>
We have been given two equations which are as follows:
3x - 4y = -24 ----- eqn 1
x + 4y = 8 -------- eqn 2
We have been asked to solve the equations which means we have to find the value of ‘x’ and ‘y’.
We rearrange eqn 2 as follows:
x + 4y = 8
x = 8 - 4y ------eqn 3
Now we substitute eqn 3 in eqn 1 as follows:
3(8 - 4y) -4y = -24
24 - 12y - 4y = -24
-16y = -48
y = 3
Substitute "y" value in eqn 3. Therefore the value of ‘x’ becomes:
x = 8 - 4(3)
x = 8 - 12 = -4
Hence on solving both the given equations we get the value of x and y as -4 and 3 respectively.
The answer to the equation is 15
The top 15% (85th percentile) is the cutoff value

such that

where
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is the corresponding cutoff for the standardized normal distribution. We have

, and so