Answer:
Approximate of error = 11.11 % (Approx.)
Step-by-step explanation:
Given:
Exact value = 50
Approximate value = 45
Find:
Approximate of error
Computation:
Approximate of error = [(Exact value - Approximate value)/Approximate value]100
Approximate of error = [(50 - 45)/45]100
Approximate of error = [(5)/45]100
Approximate of error = [0.11111]100
Approximate of error = 11.11 % (Approx.)
For this case we must find the product of the following expression:

We combine using the product rule for radicals:
![\sqrt [n] {a} * \sqrt [n] {b} = \sqrt [n] {ab}](https://tex.z-dn.net/?f=%5Csqrt%20%5Bn%5D%20%7Ba%7D%20%2A%20%5Csqrt%20%5Bn%5D%20%7Bb%7D%20%3D%20%5Csqrt%20%5Bn%5D%20%7Bab%7D)
So, we have:

We rewrite the 216 as

By definition of properties of powers and roots we have:
![\sqrt [n] {a ^ n} = a ^ {\frac {n} {n}} = a](https://tex.z-dn.net/?f=%5Csqrt%20%5Bn%5D%20%7Ba%20%5E%20n%7D%20%3D%20a%20%5E%20%7B%5Cfrac%20%7Bn%7D%20%7Bn%7D%7D%20%3D%20a)
Then, the expression is:

Answer:
Option D
Correct question with correct options has been attached
Answer:
37
Step-by-step explanation:
Let C represent those dressed as heroes and let B represent those wearing mask
Thus;
n(C) = 39
n(D) = 45
We are told that 21 entrants were dressed as heroes wearing a mask.
Thus; n(C n D) = 21
We are also told that there were a total of 100 entries.
Thus, n(T) = 100
Now, number of entrants neither dressed as heroes nor wearing a mask is given by;
n(T) - n(C u D)
n(C u D) = n(C) + n(D) - n(C n D)
n(C u D) = 39 + 45 - 21
n(C u D) = 63
Thus;number of entrants neither dressed as heroes nor wearing a mask is: 100 - 63 = 37
Well, it depends how much is each candy bar?
You know how decimals are really long to write, like 0.00000000089? Well, with scientific notation, you can shorten them. Take the decimal I wrote for instance. Using scientific notation, I can change it to 8.9 * 10^-10. The decimal is shortened, and according to the exponent, 10^-10, we move the decimal place for 8.9 to the left ten places, leaving us with 0.00000000089.