For this case we have to, by defining properties of powers and roots the following is fulfilled:
![\sqrt [n] {a ^ m} = a ^ {\frac {m} {n}}](https://tex.z-dn.net/?f=%5Csqrt%20%5Bn%5D%20%7Ba%20%5E%20m%7D%20%3D%20a%20%5E%20%7B%5Cfrac%20%7Bm%7D%20%7Bn%7D%7D)
We must rewrite the following expression:
![\sqrt [3] {8 ^ {\frac {1} {4} x}}](https://tex.z-dn.net/?f=%5Csqrt%20%5B3%5D%20%7B8%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B4%7D%20x%7D%7D)
Applying the property listed we have:
![\sqrt [3] {8 ^ {\frac {1} {4} x}} = 8 ^ {\frac{\frac {1} {4} x} {3} }= 8 ^ {\frac {1} {4 * 3} x} = 8 ^ {\frac {1} {12} x}](https://tex.z-dn.net/?f=%5Csqrt%20%5B3%5D%20%7B8%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B4%7D%20x%7D%7D%20%3D%208%20%5E%20%7B%5Cfrac%7B%5Cfrac%20%7B1%7D%20%7B4%7D%20x%7D%20%7B3%7D%20%7D%3D%208%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B4%20%2A%203%7D%20x%7D%20%3D%208%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B12%7D%20x%7D)
Using the property again we have to:
![8 ^ {\frac {1} {12} x} = \sqrt [12] {8 ^ x}](https://tex.z-dn.net/?f=8%20%5E%20%7B%5Cfrac%20%7B1%7D%20%7B12%7D%20x%7D%20%3D%20%5Csqrt%20%5B12%5D%20%7B8%20%5E%20x%7D)
Thus, the correct option is option C
Answer:
Option C
Answer:
x = -52
Step-by-step explanation:
Let us solve for x.
Use cross multiplication to find the value of x.
Divide both sides by 3.
x = - 52
Answer:
The probability that the number of free throws he makes exceeds 80 is approximately 0.50
Step-by-step explanation:
According to the given data we have the following:
P(Make a Throw) = 0.80%
n=100
Binomial distribution:
mean: np = 0.80*100= 80
hence, standard deviation=√np(1-p)=√80*0.20=4
Therefore, to calculate the probability that the number of free throws he makes exceeds 80 we would have to make the following calculation:
P(X>80)= 1- P(X<80)
You could calculate this value via a normal distributionapproximation:
P(Z<(80-80)/4)=1-P(Z<0)=1-50=0.50
The probability that the number of free throws he makes exceeds 80 is approximately 0.50
