![\displaystyle\lim_{n\to\infty}\sqrt[n]{\left|\left(\frac{5n+15}{2n-1}\right)^n\right|}=\lim_{n\to\infty}\frac{5n+15}{2n-1}=\dfrac52](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Csqrt%5Bn%5D%7B%5Cleft%7C%5Cleft%28%5Cfrac%7B5n%2B15%7D%7B2n-1%7D%5Cright%29%5En%5Cright%7C%7D%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cfrac%7B5n%2B15%7D%7B2n-1%7D%3D%5Cdfrac52)
Since this limit exceeds 1, the series diverges.
Answer:
B. 1 5/6
Step-by-step explanation:
The probability that the train will be there when Alex arrives is 5/18
If Alex arrives at any time after 1.20pm the chances that train will be there is 1/3.
However if alex arrives at 1.00pm exactly there is no chance the train will be arrive there.
The probability that the train will be there increase linearly to 1/3 as alex's arrival time moves from 1.00pm to 1.20pm.
By arranging the probabilities over the first 20 minutes to get a 1/6 chance the train will be there if alex arrives between 1.00pm to 1.20pm
we get the final answer by
=1/3( 1/6 + 1/3 + 1/3)
=5/18
So, the probability that the train will be there when Alex arrives is 5/18
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Hello there!
We can use the following expression:
$426 ÷ $48 = d
Hope this helps! Tell me if I'm wrong!
~DL
50 yards for 1 lap
50 yards x 18 laps = 900 total yards
now divide the amount he swam by the total amount he wants to swim:
738 / 900 = 0.82 = 82%
Answer is B
