Let us compute first the probability of ending up an odd number when rolling a dice. A dice has faces with numbers 1 up to 6. The odd numbers within that is 3 (1, 3 and 5). Therefore, each dice has a probability of 3/6 or 1/2. Then, you use the repeated trials formula:
Probability = n!/r!(n-r)! * p^r * q^(n-r), where n is the number of tries (n=6), r is the number tries where you get an even number (r=0), p is the probability of having an even face and q is the probability of having an odd face.
Probability = 6!/0!(6!) * (1/2)^0 * (1/2)^6
Probability = 1/64
Therefore, the probability is 1/64 or 1.56%.
Answer:
the answer is A. 3x-2+2/x+1
Answer:
10* 299.99=2999.9
2999.9/100=29.999
this would round up to 30, so Brittany saved $30. : )
Answer:
See below for proof.
Step-by-step explanation:
<u>Given</u>:

<u>First derivative</u>

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<u>Second derivative</u>
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<u>Proof</u>



![= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]](https://tex.z-dn.net/?f=%3D%20%5Cleft%28x%2B%5Csqrt%7B1%2Bx%5E2%7D%5Cright%29%5Em%5Cleft%5Bm%5E2-%5Cdfrac%7Bmx%7D%7B%5Csqrt%7B1%2Bx%5E2%7D%7D%2B%5Cdfrac%7Bmx%7D%7B%5Csqrt%7B1%2Bx%5E2%7D%7D-m%5E2%5Cright%5D)
![= \left(x+\sqrt{1+x^2}\right)^m\left[0]](https://tex.z-dn.net/?f=%3D%20%5Cleft%28x%2B%5Csqrt%7B1%2Bx%5E2%7D%5Cright%29%5Em%5Cleft%5B0%5D)

Whats not affected by the
others outcome