Answer:
![\frac{\sqrt[3]{16y^4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B16y%5E4%7D%7D%7Bx%5E2%7D)
Step-by-step explanation:
The options are missing; However, I'll simplify the given expression.
Given
![\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B32x%5E3y%5E6%7D%7D%7B%5Csqrt%5B3%5D%7B2x%5E9y%5E2%7D%20%7D)
Required
Write Equivalent Expression
To solve this expression, we'll make use of laws of indices throughout.
From laws of indices ![\sqrt[n]{a} = a^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D%20%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
So,
gives
Also from laws of indices
So, the above expression can be further simplified to
Multiply the exponents gives
Substitute 
for 32
From laws of indices
This law can be applied to the expression above;
becomes
Solve exponents
From laws of indices,
; So,
gives
The expression at the numerator can be combined to give
Lastly, From laws of indices,
![a^{\frac{m}{n} = \sqrt[n]{a^m}](https://tex.z-dn.net/?f=a%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%20%3D%20%5Csqrt%5Bn%5D%7Ba%5Em%7D)
; So,
becomes
![\frac{\sqrt[3]{(2y)}^{4}}{x^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B%282y%29%7D%5E%7B4%7D%7D%7Bx%5E2%7D)
Hence,
is equivalent to
Answer:
a) 0.05
b) 0.9826
c) 0.000039308
Step-by-step explanation:
a) 
b) For two minutes, the mean is doubled, hence it is 6. In order to calculate the probability of al least two calls arriving, we calculate first the probability of the complementary event: At most 1 call will arrive. For that probability, we need to sum the probabilities of 0 and 1.
Hence,
c) For five minutes the mean is 15. We need to sum the probabilities of 0, 1 and 2.
As a result,
Practically 0
Answer: 2^60 = 8^20
Step-by-step explanation: 8=2^3
(2^3)^20 Multiply the exponents
Answer:
Option C: 0.28
Step-by-step explanation:
This is a binomial probability distribution problem.
Now, we want to find the probability that at least 2 thumbtacks land pointing up when 5 thumbtacks are tossed. This is written as;
P(X ≥ 2) = P(2) + P(3) + P(4) + P(5)
From the histogram;
P(5) = 0.02
P(4) = 0.02
P(3) = 0.05
P(2) = 0.19
Thus;
P(X ≥ 2) = 0.19 + 0.05 + 0.02 + 0.02
P(X ≥ 2) = 0.28
Answer:
No
Step-by-step explanation:
The first nine in on the place of the thousands, while the second nine is on the place of the tens. so the first nine is a hundred times as great as the second nine
