That is 6120.43 yen rounded to the nearest hundredth because if you exchange 56.92 at a rate of 1 yen is equal to 0.0093 you get 6120.43.
0.433 is already rounded to the hundredths so to round it to the nearest tenths, remove the last number.
Your answer is 0.43
Answer:
1. 7/41
2. 7/15
Step-by-step explanation:
1. Probability that it is a multiple of 6 given that it is a 2 digit number: There are 7 numbers that are a multiple of 6 and are 2 digits: 12, 18, 24, 30, 36, 42, and 48. There are 41 numbers that are 2 digits. So, it is 7/41
2. Probability that it is at least 20 given that it is prime. Out of the prime numbers from 1-50, 7 of them are at least 20 and there are 15 primes in that range in total. So, it is 7/15
From the right hand side, we will need to find a way to rewriting 3x²y in terms of cube roots.
We know that 27 is 3³, so if we were to rewrite it in terms of cube roots, we will need to multiply everything by itself two more twice. (ie we can rewrite it as ∛(3x²y)³)
Hence, we can say that it's:
![\sqrt[3]{162x^{c}y^{5}} = \sqrt[3]{(3x^{2}y)^{3}} * \sqrt[3]{6y^{d}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B162x%5E%7Bc%7Dy%5E%7B5%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B%283x%5E%7B2%7Dy%29%5E%7B3%7D%7D%20%2A%20%5Csqrt%5B3%5D%7B6y%5E%7Bd%7D%7D)
![= \sqrt[3]{162x^{6}y^{3+d}}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%5B3%5D%7B162x%5E%7B6%7Dy%5E%7B3%2Bd%7D%7D)
Hence, c = 6 and d = 2