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Aleks [24]
3 years ago
12

The odds of flipping a fair coin either heads or tails is 50%. The odds of rolling a six on a fair six sided die is 16.6% or 1 i

n 6. Let’s suppose that we have three regular six sided dice. If you roll all three dice at the same time, what are the odds that the sum of the numbers that show up on you dice are a prime number?
Mathematics
1 answer:
creativ13 [48]3 years ago
8 0

Answer:

P( sum is prime )= 73/216

Step-by-step explanation:

The minimum value of the sum will be 3 and maximum value will be 18. So the prime numbers in this range are 3 , 5, 7, 11, 13, 17.

P(sum=3)=1/216, P(sum=5)=6/216, P(sum=7)=15/216, P(sum=11)=27/216, P(sum=13)=21/216, P(sum=17)=3/216.

The final probability will be sum of the above given probabilities.

Hence P( sum is prime )= 73/216

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Consider functions of the form f(x)=a^x for various values of a. In particular, choose a sequence of values of a that converges
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A. As "a"⇒e, the function f(x)=aˣ tends to be its derivative.

Step-by-step explanation:

A. To show the stretched relation between the fact that "a"⇒e and the derivatives of the function, let´s differentiate f(x) without a value for "a" (leaving it as a constant):

f(x)=a^{x}\\ f'(x)=a^xln(a)

The process will help us to understand what is happening, at first we rewrite the function:

f(x)=a^x\\ f(x)=e^{ln(a^x)}\\ f(x)=e^{xln(a)}\\

And then, we use the chain rule to differentiate:

f'(x)=e^{xln(a)}ln(a)\\ f'(x)=a^xln(a)

Notice the only difference between f(x) and its derivative is the new factor ln(a). But we know  that ln(e)=1, this tell us that as "a"⇒e, ln(a)⇒1 (because ln(x) is a continuous function in (0,∞) ) and as a consequence f'(x)⇒f(x).

In the graph that is attached it´s shown that the functions follows this inequality (the segmented lines are the derivatives):

if a<e<b, then aˣln(a) < aˣ < eˣ < bˣ < bˣln(b)  (and below we explain why this happen)

Considering that ln(a) is a growing function and ln(e)=1, we have:

if a<e<b, then ln(a)< 1 <ln(b)

if a<e, then aˣln(a)<aˣ

if e<b, then bˣ<bˣln(b)

And because eˣ is defined to be the same as its derivative, the cases above results in the following

if a<e<b, then aˣ < eˣ < bˣ (because this function is also a growing function as "a" and "b" gets closer to e)

if a<e, then aˣln(a)<aˣ<eˣ ( f'(x)<f(x) )

if e<b, then eˣ<bˣ<bˣln(b) ( f(x)<f'(x) )

but as "a"⇒e, the difference between f(x) and f'(x) begin to decrease until it gets zero (when a=e)

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