Answer: ![3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Work Shown:
![\sqrt[3]{27x^{6}y^{4}}\\\\\sqrt[3]{3^3x^{3+3}y^{3+1}}\\\\\sqrt[3]{3^3x^{3}*x^{3}*y^{3}*y^{1}}\\\\\sqrt[3]{3^3x^{2*3}*y^{3}*y}\\\\\sqrt[3]{\left(3x^2y\right)^3*y}\\\\\sqrt[3]{\left(3x^2y\right)^3}*\sqrt[3]{y}\\\\3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27x%5E%7B6%7Dy%5E%7B4%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%2B3%7Dy%5E%7B3%2B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%7D%2Ax%5E%7B3%7D%2Ay%5E%7B3%7D%2Ay%5E%7B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B2%2A3%7D%2Ay%5E%7B3%7D%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%7D%2A%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Explanation:
As the steps above show, the goal is to factor the expression under the root in terms of pulling out cubed terms. That way when we apply the cube root to them, the exponents cancel. We cannot factor the y term completely, so we have a bit of leftovers.
Answer:
53 cupcakes
Step-by-step explanation:
Take the amount of frosting and divide by the amount needed per cupcake
431/8
53.875
We round down since we cannot frost part of a cupcake
53 cupcakes
Whatever the number of times you flip the disc, as long as the disc has uniformly distributed mass throughout the body, then
The probability will always be 50%
Answer:
Given the mean = 205 cm and standard deviation as 7.8cm
a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.
b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.
c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.
Step-by-step explanation:
Given the mean = 205 cm and standard deviation as 7.8cm
a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.
b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.
c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.
Step-by-step explanation:
it is 6 and 7 so that's because Tom Holland is baby daddy