Answer:
The answer is the option C
cube root of 
Step-by-step explanation:
Remember that
![a^{\frac{x}{y}} =\sqrt[y]{a^{x}}](https://tex.z-dn.net/?f=a%5E%7B%5Cfrac%7Bx%7D%7By%7D%7D%20%3D%5Csqrt%5By%5D%7Ba%5E%7Bx%7D%7D)
in this problem we have
therefore
![2^{\frac{4}{3}} =\sqrt[3]{2^{4}}=\sqrt[3]{16}](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B4%7D%7B3%7D%7D%20%3D%5Csqrt%5B3%5D%7B2%5E%7B4%7D%7D%3D%5Csqrt%5B3%5D%7B16%7D)
You have the following function:
g(x) = 3x - 7
In order to determine th value of x when g(x) = -4, you equal the previous expression to -4 and solve for x, just as follow:
g(x) = 3x - 7 replace g(x) = -4
-4 = 3x - 7 add 7 both sides
-4 + 7 = 3x simplify
3 = 3x divide by 3 both sides
3/3 = x
x = 1
Hence, the anser is x = 1
Answer:
30/552
Step-by-step explanation:
In order to solve this problem you need to multiply the probability of getting grape for the first gumball with the probability of getting grape for the second gumball. Since there are 6 grape gumballs and a total of 24 gumballs (6*4). Then the probability of getting grape for the firs one is
Now there are only 5 grape gumballs available and one less in the total supply, therefore the probability of getting grape in the second try is
Finally we multiply them together to find the probability of getting two grapes in a row.
= 
Answer:
D.- ac
Step-by-step explanation:
mark me as brainliest pls
Supposing, for the sake of illustration, that the mean is 31.2 and the std. dev. is 1.9.
This probability can be calculated by finding z-scores and their corresponding areas under the std. normal curve.
34 in - 31.2 in
The area under this curve to the left of z = -------------------- = 1.47 (for 34 in)
1.9
32 in - 31.2 in
and that to the left of 32 in is z = ---------------------- = 0.421
1.9
Know how to use a table of z-scores to find these two areas? If not, let me know and I'll go over that with you.
My TI-83 calculator provided the following result:
normalcdf(32, 34, 31.2, 1.9) = 0.267 (answer to this sample problem)
