Answer:
340
Step-by-step explanation: 200 * .14 = 28 then 28*5= 140 then 140+200=340
Answer:
D
Step-by-step explanation:
There are 2 cups in a pint, so 3 cups would be 1.5 pints. :)
Based on the data, the median time it takes players to complete the game is 1148 minutes.
<h3>What is the median?</h3>
It is the mathematical value that is the middle when values are sorted.
<h3>How to find the median time?</h3>
457 - 548- 866 - 952 - 976 - 1037 - 1148 - 1235 - 1245- 1431-1486 - 1759 - 1864
Identify the value of the middle
Considering there are 13 values, the value of the middle is the value number 7 or 1148.
Note: This question is incomplete because the data is missing, below I attached the missing section,
Learn more about median time in: brainly.com/question/7493103
68% because it’s between them
Megan:
x to the one third power =

<span>x to the one twelfth power = </span>

<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>

<span>
Since </span>

then

Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4

Julie:
x times x to the second times x to the fifth = x * x² * x⁵
<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>
![\sqrt[32]{x* x^{2} * x^{5} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B32%5D%7Bx%2A%20x%5E%7B2%7D%20%2A%20x%5E%7B5%7D%20%7D%20)
<span>
Since </span>

Then
![\sqrt[32]{x* x^{2} * x^{5} }= \sqrt[32]{ x^{1+2+5} } =\sqrt[32]{ x^{8} }](https://tex.z-dn.net/?f=%5Csqrt%5B32%5D%7Bx%2A%20x%5E%7B2%7D%20%2A%20x%5E%7B5%7D%20%7D%3D%20%5Csqrt%5B32%5D%7B%20x%5E%7B1%2B2%2B5%7D%20%7D%20%3D%5Csqrt%5B32%5D%7B%20x%5E%7B8%7D%20%7D)
Since
![\sqrt[n]{x^{m}} = x^{m/n} }](https://tex.z-dn.net/?f=%20%5Csqrt%5Bn%5D%7Bx%5E%7Bm%7D%7D%20%3D%20x%5E%7Bm%2Fn%7D%20%7D%20)
Then
![\sqrt[32]{ x^{8} }= x^{8/32} = x^{1/4}](https://tex.z-dn.net/?f=%5Csqrt%5B32%5D%7B%20x%5E%7B8%7D%20%7D%3D%20x%5E%7B8%2F32%7D%20%3D%20x%5E%7B1%2F4%7D%20)
Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.