They had the same quotient because both of those equations are equal
What is the probability that you will get exactly zero
heads? What is the probability that you will get exactly one head? What is the probability that you will get exactly 4 head? If it helps, there are <span><span><span><span>2 to the </span><span>4th power... </span></span> </span><span>24</span></span>
possibilities for the sequence of four flips. Try writing them all out and see if you can spot a pattern.
<h3>
Answer:</h3>
C. 4(x-3)-x
<h3>
Step-by-step explanation:</h3>
All of the given expressions are equivalent to 3x+12 except selection C. Using that in your equation makes it be ...
... 3(x +1) +9 = 4(x -3) -x
... 3x +12 = 3x -12
... 12 = -12 . . . . . <em>false</em>
There is no value of x that will make this true, hence NO SOLUTION.
_____
<em>Comment on the other choices</em>
3x+12 = 3x+12 has an <em>infinite number of solutions</em>, as <u><em>any</em></u> value of x will make this true.
Answer:
(a) The probability of the event (<em>X</em> > 84) is 0.007.
(b) The probability of the event (<em>X</em> < 64) is 0.483.
Step-by-step explanation:
The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 64.
The probability mass function of a Poisson distribution is:
(a)
Compute the probability of the event (<em>X</em> > 84) as follows:
P (X > 84) = 1 - P (X ≤ 84)
![=1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007](https://tex.z-dn.net/?f=%3D1-%5Csum%20_%7Bx%3D0%7D%5E%7Bx%3D84%7D%5Cfrac%7Be%5E%7B-64%7D%2864%29%5E%7Bx%7D%7D%7Bx%21%7D%5C%5C%3D1-%5Be%5E%7B-64%7D%5Csum%20_%7Bx%3D0%7D%5E%7Bx%3D84%7D%5Cfrac%7B%2864%29%5E%7Bx%7D%7D%7Bx%21%7D%5D%5C%5C%3D1-%5Be%5E%7B-64%7D%5B%5Cfrac%7B%2864%29%5E%7B0%7D%7D%7B0%21%7D%2B%5Cfrac%7B%2864%29%5E%7B1%7D%7D%7B1%21%7D%2B%5Cfrac%7B%2864%29%5E%7B2%7D%7D%7B2%21%7D%2B...%2B%5Cfrac%7B%2864%29%5E%7B84%7D%7D%7B84%21%7D%5D%5D%5C%5C%3D1-0.99308%5C%5C%3D0.00692%5C%5C%5Capprox0.007)
Thus, the probability of the event (<em>X</em> > 84) is 0.007.
(b)
Compute the probability of the event (<em>X</em> < 64) as follows:
P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)
![=\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483](https://tex.z-dn.net/?f=%3D%5Csum%20_%7Bx%3D0%7D%5E%7Bx%3D63%7D%5Cfrac%7Be%5E%7B-64%7D%2864%29%5E%7Bx%7D%7D%7Bx%21%7D%5C%5C%3De%5E%7B-64%7D%5Csum%20_%7Bx%3D0%7D%5E%7Bx%3D63%7D%5Cfrac%7B%2864%29%5E%7Bx%7D%7D%7Bx%21%7D%5C%5C%3De%5E%7B-64%7D%5B%5Cfrac%7B%2864%29%5E%7B0%7D%7D%7B0%21%7D%2B%5Cfrac%7B%2864%29%5E%7B1%7D%7D%7B1%21%7D%2B%5Cfrac%7B%2864%29%5E%7B2%7D%7D%7B2%21%7D%2B...%2B%5Cfrac%7B%2864%29%5E%7B63%7D%7D%7B63%21%7D%5D%5C%5C%3D0.48338%5C%5C%5Capprox0.483)
Thus, the probability of the event (<em>X</em> < 64) is 0.483.
