Just need someone to clarify things with me!
As I was solving this equation, I got:
x^-2
-------
x^10
I moved x^-2 down with x^10
So in x^-2's place, I placed a positive 1.
1
----------------------
x^10 times x^-2
The radical rule here is to pretty much subtract ahead. However, this extremely smart person said something like if I move a negative number to the denominator, it becomes positive. However, in this case, x^-2 stays the same.
Anyone explain here??
Answers
Answer:
[tex]\frac{1}{x^{12} }[/tex]
Step-by-step explanation:
Given [tex](\frac{x}{x^{-5}})^{-2}[/tex]
The Negative Exponent Rule states that, [tex]a^{-n} = \frac{1}{a^{n}}[/tex]
Basically, since you're raising the fraction inside the parenthesis to a negative exponent, the whole fraction becomes a denominator of 1: [tex]\frac{1}{(\frac{x}{x^{-5}})^{2}}[/tex]
Then, you'll have to work on the denominator:
[tex]\frac{1}{(\frac{x}{x^{-5}})^{2}} = \frac{1}{\frac{x^{2} }{(x^{-5})^{2}}}[/tex]
Next, you'll have to work on the divisor of x² (in the denominator), [tex](x^{-5})^{2}[/tex] by using the Power-to-Power Rule of Exponents: [tex](a^{m})^{n} = a^{mn}[/tex], which results in:
[tex](x^{-5})^{2} = x^{-10}[/tex]
Since it is a negative exponent, you'll have to apply the Negative Exponent Rule once again to the denominator.
[tex]\frac{1}{(\frac{x}{x^{-5}})^{2}} = \frac{1}{\frac{x^{2} }{(x^{-5})^{2}}} = \frac{1}{\frac{x^{2} }{x^{-10}}} = \frac{1}{\frac{x^{2} }{\frac{1}{x^{10}}}}[/tex]
At this point, you could apply the Fraction rule onto the denominator: [tex]\frac{a}{\frac{b}{c}} = \frac{a*c}{b}[/tex]
So the denominator now becomes:
[tex]\frac{1}{\frac{x^{2} }{\frac{1}{x^{10}}}} = \frac{1}{x^{2}*x^{10} }[/tex]
Finally, you could apply the Product Rule of Exponents, [tex]a^{m}a^{n} = a^{m + n }[/tex] onto the denominator:
[tex]\frac{1}{x^{2}*x^{10} } = \frac{1}{x^{2+10}} = \frac{1}{x^{12} }[/tex]
Therefore, the correct answer is: [tex]\frac{1}{x^{12} }[/tex]
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Binomial probability distribution
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0.0001 probability of an accident on a given day, thus [tex]p = 0.0001[/tex]250 days, for 35 years, thus the number of days is [tex]n = 250(35) = 8750[/tex].Item a:
The probability is P(X = 0), thus:
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[tex]P(X = 0) = C_{8750,0}.(0.0001)^{0}.(0.9999)^{8750} = 0.4168[/tex]
0.4168 = 41.68% probability that he gets through his career without a serious cycling accident.
Item b:
The probability changes, as we consider now just five years.The number of days is now [tex]n = 250(5) = 1250[/tex].Then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
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Considering that the probability of an accident on a given day is independent of any other day, now just the last five years are considered, thus, the probability changes and there is a 0.8825 = 88.25% probability that he will get through his entire career without an accident the same.
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