Answer: ![x=\Large\boxed{\frac{m}{y-z} }](https://tex.z-dn.net/?f=x%3D%5CLarge%5Cboxed%7B%5Cfrac%7Bm%7D%7By-z%7D%20%7D)
Step-by-step explanation:
<u>Given expression</u>
![z=y-\dfrac{m}{x}](https://tex.z-dn.net/?f=z%3Dy-%5Cdfrac%7Bm%7D%7Bx%7D)
<u>Subtract [ y ] on both sides</u>
![z-y=y-\dfrac{m}{x} -y](https://tex.z-dn.net/?f=z-y%3Dy-%5Cdfrac%7Bm%7D%7Bx%7D%20-y)
![z-y=-\dfrac{m}{x}](https://tex.z-dn.net/?f=z-y%3D-%5Cdfrac%7Bm%7D%7Bx%7D)
<u>Multiply [ -x ] on both sides</u>
![(-x)(z-y)=(-x)(-\dfrac{m}{x} )](https://tex.z-dn.net/?f=%28-x%29%28z-y%29%3D%28-x%29%28-%5Cdfrac%7Bm%7D%7Bx%7D%20%29)
![(x)(-1)(z-y)=m](https://tex.z-dn.net/?f=%28x%29%28-1%29%28z-y%29%3Dm)
![(x)(y-z)=m](https://tex.z-dn.net/?f=%28x%29%28y-z%29%3Dm)
<u>Divide [ y - z ] on both sides</u>
![(x)(y-z)\div(y-z)=m\div(y-z)](https://tex.z-dn.net/?f=%28x%29%28y-z%29%5Cdiv%28y-z%29%3Dm%5Cdiv%28y-z%29)
![x=\Large\boxed{\frac{m}{y-z} }](https://tex.z-dn.net/?f=x%3D%5CLarge%5Cboxed%7B%5Cfrac%7Bm%7D%7By-z%7D%20%7D)
Hope this helps!! :)
Please let me know if you have any questions
Answer:
560
Step-by-step explanation:
this is solved above, I hope it helps
Answer:
the ball is on the ground after 3 seconds right so it would still be on the ground
Step-by-step explanation:
<h3><u>Write 3 fractions, using 10 as the denominator, who's sum is 1</u></h3>
![\frac{1}{10}, \frac{3}{10}, \frac{6}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B10%7D%2C%20%5Cfrac%7B3%7D%7B10%7D%2C%20%5Cfrac%7B6%7D%7B10%7D)
<em><u>Solution:</u></em>
Given that,
We have to write 3 fractions, using 10 as the denominator, who's sum is 1
We know that, fraction is written as:
![fraction = \frac{numerator}{denominator}](https://tex.z-dn.net/?f=fraction%20%3D%20%5Cfrac%7Bnumerator%7D%7Bdenominator%7D)
We have to write 3 fractions with 10 as denominator
![\frac{1}{10}\\\\\frac{3}{10}\\\\\frac{6}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B10%7D%5C%5C%5C%5C%5Cfrac%7B3%7D%7B10%7D%5C%5C%5C%5C%5Cfrac%7B6%7D%7B10%7D)
Add these to check if its sum is 1
![sum = \frac{1}{10} + \frac{3}{10} + \frac{6}{10}](https://tex.z-dn.net/?f=sum%20%3D%20%5Cfrac%7B1%7D%7B10%7D%20%2B%20%5Cfrac%7B3%7D%7B10%7D%20%2B%20%5Cfrac%7B6%7D%7B10%7D)
When denominators are same, we can add the numerators
![sum = \frac{1+3+6}{10}\\\\sum = \frac{10}{10}\\\\sum = 1](https://tex.z-dn.net/?f=sum%20%3D%20%5Cfrac%7B1%2B3%2B6%7D%7B10%7D%5C%5C%5C%5Csum%20%3D%20%5Cfrac%7B10%7D%7B10%7D%5C%5C%5C%5Csum%20%3D%201)
Thus the three fractions are:
![\frac{1}{10}, \frac{3}{10}, \frac{6}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B10%7D%2C%20%5Cfrac%7B3%7D%7B10%7D%2C%20%5Cfrac%7B6%7D%7B10%7D)
Answer:
The median of the distribution is (first option);
1
Step-by-step explanation:
The values on the given table are presented as follows;
![\begin{array}{cccccc}Value \ of \ Spin&0&1&2&5&10\\Probability&0.4&0.2&0.2&0.1&0.1\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bcccccc%7DValue%20%5C%20of%20%5C%20Spin%260%261%262%265%2610%5C%5CProbability%260.4%260.2%260.2%260.1%260.1%5Cend%7Barray%7D)
To find the median, we calculate the cumulative probabilities to find the value of the spin art which the cumulative probability is 0.5 as follows;
Cumulative probability, starting from the left, is therefore;
Probability for spin of 0, 0.4 + Probability for spin of 1, 0.2 = 0.6
Therefore, given that the probability of 0.5 occurs within the Value of Spin 1, the median of the probability distribution is 1