Answer: 
Step-by-step explanation:
In order to round to the nearest thousandth, you must observe the digit to the right of the digit in the thousandth place and then:
- If it is less than 5, you must round down.
- If it is greater than or equal than 5, you must round up.
In this case, given the number:
You can notice that the digit to the right of the digit in the thousandth place is 9.
Since 
, you must round up. Then:
Finally, to express this as a single-digit times a power of ten, move the decimal point three places to the right. Then you get:
<h3>
Answer: Choice A</h3>
![x^2\left(\sqrt[4]{x^2}\right)](https://tex.z-dn.net/?f=x%5E2%5Cleft%28%5Csqrt%5B4%5D%7Bx%5E2%7D%5Cright%29)
=====================================================
Explanation:
The fourth root of x is the same as x^(1/4)
I.e,
![\sqrt[4]{x} = x^{1/4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%7D%20%3D%20x%5E%7B1%2F4%7D)
The same applies to x^10 as well
![\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%20%3D%20%5Cleft%28x%5E%7B10%7D%5Cright%29%5E%7B1%2F4%7D)
Multiply the exponents 10 and 1/4 to get 10/4
![\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4} = x^{10*1/4} = x^{10/4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%20%3D%20%5Cleft%28x%5E%7B10%7D%5Cright%29%5E%7B1%2F4%7D%20%3D%20x%5E%7B10%2A1%2F4%7D%20%3D%20x%5E%7B10%2F4%7D)
![\sqrt[4]{x^{10}} = x^{10/4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%20%3D%20x%5E%7B10%2F4%7D)
-----------------------
If we have an expression in the form x^(m/n), with m > n, then we can simplify it into an equivalent form as shown below
![x^{m/n} = x^a\sqrt[n]{x^b}](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20x%5Ea%5Csqrt%5Bn%5D%7Bx%5Eb%7D)
The 'a' and 'b' are found through dividing m/n
m/n = a remainder b
'a' is the quotient, b is the remainder
-----------------------
The general formula can easily be confusing, so let's replace m and n with the proper numbers. In this case, m = 10 and n = 4
m/n = 10/4 = 2 remainder 2
We have a = 2 and b = 2
So
turns into
![x^{10/4} = x^2\sqrt[4]{x^2}](https://tex.z-dn.net/?f=x%5E%7B10%2F4%7D%20%3D%20x%5E2%5Csqrt%5B4%5D%7Bx%5E2%7D)
which means
![\sqrt[4]{x^{10}} = {x^2} \sqrt[4]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E%7B10%7D%7D%20%3D%20%7Bx%5E2%7D%20%5Csqrt%5B4%5D%7Bx%5E2%7D)
<span>the answer issssssssssss, 3713000</span>
Answer:
b
Step-by-step explanation:
2022 on edge I just did it
We have to calculate the probability of picking a 4 and then a 5 without replacement.
We can express this as the product of the probabilities of two events:
• The probability of picking a 4
,
• The probability of picking a 5, given that a 4 has been retired from the deck.
We have one card in the deck out of fouor cards that is a "4".
Then, the probability of picking a "4" will be:
The probability of picking a "5" will be now equal to one card (the number of 5's in the deck) divided by the number of remaining cards (3 cards):
We then calculate the probabilities of this two events happening in sequence as:
Answer: 1/12
