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.
0 divided by 2 is 0.
You can use the multiplicative identity property: the product of any number and zero is still zero.
Hope this helped :)
Answer:
<u>y'= 5x^4 + 5^x In(5)</u>
Step-by-step explanation:
<u>Differentiate</u><u> </u><u>with </u><u>Respect</u><u> </u><u>to</u><u> </u><u>x</u>
<u>f(</u><u>x)</u><u>'</u><u>=</u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>In(</u><u>5</u><u>^</u><u>x</u><u>)</u>
<u>f(</u><u>x)</u><u>'</u><u>=</u><u> </u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>x </u><u>In(</u><u>5</u><u>)</u>
<u>with </u><u>respect</u><u> </u><u>to </u><u>x,</u><u> </u><u>we </u><u>have</u>
<u>y'=</u><u> </u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>y </u><u>In(</u><u>5</u><u>)</u>
<u>y'=</u><u> </u><u>5</u><u>x</u><u>^</u><u>4</u><u> </u><u>+</u><u> </u><u>5</u><u>^</u><u>x</u><u> </u><u>In(</u><u>5</u><u>)</u>
Answer:
1/4 hopefully this helps you with work