Answer:
C
Step-by-step explanation:
The vertical line test is basically just drawing a vertical line and seeing if the line intersects the graph more than once. If it does, then it is not a function, if it doesn't than it is a function.
Answer:
i do no sorry plz i have exam in online so sorry
Answer:
A
Step-by-step explanation:
Answer:
1. ![(\sqrt[5]{(m+2)})^{3} = (m+2)^{\frac{3}{5}}](https://tex.z-dn.net/?f=%28%5Csqrt%5B5%5D%7B%28m%2B2%29%7D%29%5E%7B3%7D%20%3D%20%20%28m%2B2%29%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D)
2. ![(\sqrt[3]{(m+2)})^{5} = (m+2)^{\frac{5}{3}}](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7B%28m%2B2%29%7D%29%5E%7B5%7D%20%3D%20%20%28m%2B2%29%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D)
3. ![\sqrt[5]{(m)}^{3}+2 = m^{\frac{3}{5}}+2](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B%28m%29%7D%5E%7B3%7D%2B2%20%3D%20%20m%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D%2B2)
4. ![\sqrt[3]{(m)}^{5}+2 = m^{\frac{5}{3}}+2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%28m%29%7D%5E%7B5%7D%2B2%20%3D%20%20m%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%2B2)
Step-by-step explanation:
Recall that
![(\sqrt[n]{x})^{m} = (x^{\frac{m}{n}})](https://tex.z-dn.net/?f=%28%5Csqrt%5Bn%5D%7Bx%7D%29%5E%7Bm%7D%20%3D%20%20%28x%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%29)
Where
is called radicand and n is called index
1. Root(5, (m + 2) ^ 3)
In this case,
n is 5
m is 3
x = (m + 2)
![(\sqrt[5]{(m+2)})^{3} = (m+2)^{\frac{3}{5}}](https://tex.z-dn.net/?f=%28%5Csqrt%5B5%5D%7B%28m%2B2%29%7D%29%5E%7B3%7D%20%3D%20%20%28m%2B2%29%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D)
2. Root(3, (m + 2) ^ 5)
In this case,
n is 3
m is 5
x = (m + 2)
![(\sqrt[3]{(m+2)})^{5} = (m+2)^{\frac{5}{3}}](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7B%28m%2B2%29%7D%29%5E%7B5%7D%20%3D%20%20%28m%2B2%29%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D)
3. Root(5, m ^ 3) + 2
In this case,
n is 5
m is 3
x = m
![\sqrt[5]{(m)}^{3}+2 = m^{\frac{3}{5}}+2](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B%28m%29%7D%5E%7B3%7D%2B2%20%3D%20%20m%5E%7B%5Cfrac%7B3%7D%7B5%7D%7D%2B2)
4. Root(3, m ^ 5) + 2
In this case,
n is 3
m is 5
x = m
![\sqrt[3]{(m)}^{5}+2 = m^{\frac{5}{3}}+2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%28m%29%7D%5E%7B5%7D%2B2%20%3D%20%20m%5E%7B%5Cfrac%7B5%7D%7B3%7D%7D%2B2)
This is an example of conditional probability because we are trying to find the probability of an event occurring GIVEN the occurrence of some other event. There is a formula for this (see image attached).
If we follow this formula, the numerator would be the probability of (A AND B) which in this case is "48% of the class passed BOTH exams." The denominator in the formula would be that "60% of the class passed ONLY THE SECOND exam."
Therefore, P(A and B) = 0.48, which is 48% expressed as a decimal and P(B)= 0.60, which is 60% expressed as a decimal. Then, you can figure out the answer by dividing.