Isaac and William, in the mall parking lot, have found 11 quarters, 3 fifty-cent pieces, 36 dimes, 40 nickels, and 134 pennies.
2 answers:
Answer:
Step-by-step explanation:
The experimental probability is based on what already happened.
So we see that both of them found a total of 11 + 3 + 36 + 40 + 134 = 224 coins.
The coins that are worth MORE THAN 10 cents are 11 quarters and 3 fifty-cent pieces (we exclude dimes because we want MORE THAN 10 cents, NOT 10 cents exactly).
So 11 + 3 = 14 coins are worth MORE THAN 10.
So the probability of finding one worth more than 10 cents next is 14/224 = 1/16
You might be interested in
Usually one will differentiate the function to find the minimum/maximum point, but in this case differentiating yields:
which contains multiple solution if one tries to solve for x when the differentiated form is 0.
I would, though, venture a guess that the minimum value would be (approaching) 5, since the function would be undefined in the vicinity.
If, however, the function is
Then differentiating and equating to 0 yields:
which gives:
or 
We reject x=5 as it is when it ix the maximum and thus,
, for 
which contains multiple solution if one tries to solve for x when the differentiated form is 0.
I would, though, venture a guess that the minimum value would be (approaching) 5, since the function would be undefined in the vicinity.
If, however, the function is
Then differentiating and equating to 0 yields:
which gives:
We reject x=5 as it is when it ix the maximum and thus,
Remember, order of operations
exponent before multiply
so 4x^5/6=4 times x^5/6
simplify the x^5/6 first
remember
![x^\frac{m}{n}=\sqrt[n]{s^m}](https://tex.z-dn.net/?f=x%5E%5Cfrac%7Bm%7D%7Bn%7D%3D%5Csqrt%5Bn%5D%7Bs%5Em%7D)
so
![x^\frac{5}{6}=\sqrt[6]{x^5}](https://tex.z-dn.net/?f=x%5E%5Cfrac%7B5%7D%7B6%7D%3D%5Csqrt%5B6%5D%7Bx%5E5%7D)
so
![4x^\frac{5}{6}=4\sqrt[6]{x^5}](https://tex.z-dn.net/?f=4x%5E%5Cfrac%7B5%7D%7B6%7D%3D4%5Csqrt%5B6%5D%7Bx%5E5%7D)
<span />
exponent before multiply
so 4x^5/6=4 times x^5/6
simplify the x^5/6 first
remember
so
so
<span />