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yulyashka [42]
3 years ago
5

How to do that? (Indices) a

{1}{2} " alt=" \frac{1}{2} " align="absmiddle" class="latex-formula">(a+\frac{1}{a})
Mathematics
1 answer:
Tamiku [17]3 years ago
3 0
\bf a\cdot \cfrac{1}{2}\left(a+\cfrac{1}{a}  \right)\impliedby \textit{let's distribute first}
\\\\\\
\cfrac{a}{2}\left(a+\cfrac{1}{a}  \right)\implies \cfrac{a}{2}\cdot a+\cfrac{\underline{a}}{2}\cdot \cfrac{1}{\underline{a}}\implies \cfrac{a^2}{2}+\cfrac{1}{2}\implies \cfrac{a^2+1}{2}
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According to a study conducted in one city, 35% of adults in the city have credit card debts more than $2.000. A simple random s
BabaBlast [244]

Answer:

Binomial; \mu p=87.5, \sigma p=7.542

Step-by-step explanation:

  • a distribution is said be a binomial distribution iff
  1. The probability of success of that event( let it be p) is same for every trial
  2. each trial should have 2 outcome : p or (1-p) i.e, success or failure only.
  3. there are fixed number of trials (n)
  4. the trials are independent
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  • here n=250
  • the probability of success(0.35) is same for every trial

(35/100=0.35 is the required p here)

  • \mu_{p} =n*p=250*\frac{35}{100} =250*0.35=87.5

[since, the formula for \mu _{p} =n*p ]

  • \sigma _{p} =\sqrt{n*(p)*(1-p)} = \sqrt{250*0.35*(1-0.35)} = 7.542 ( approximately)

[since, the formula for [tex]\sigma _{p} =\sqrt{n*(p)*(1-p)}

  • therefore, it is Binomial; \mu p=87.5, \sigma p=7.542

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