A=1/2h(a+b), solve for a
2 answers:
You might be interested in
When you roll a number cube, there is a possibility of a number from 1 to 6 appearing. i.e. 1, 2, 3, 4, 5, or 6 can appear.
The same goes for the second number cube.
The table below presents the possible outcomes of rolling two number cubes with the sum written as exponent.
![\begin{center} \begin{tabular} {| c || c | c | c | c | c | c |} & 1 & 2 & 3 & 4 & 5 & 6 \\ [1ex] 1 & \{1,1\}^2 & \{1,2\}^3 & \{1,3\}^4 & \{1,4\}^5 & \{1,5\}^6 & \{1,6\}^7 \\ 2 & \{2,1\}^3 & \{2,2\}^4 & \{2,3\}^5 & \{2,4\}^6 & \{2,5\}^7 & \{2,6\}^8 \\ 3& \{3,1\}^4 & \{3,2\}^5 & \{3,3\}^6 & \{3,4\}^7 & \{3,5\}^8 & \{3,6\}^9 \\ 4 & \{4,1\}^5 & \{4,2\}^6 & \{4,3\}^7 & \{4,4\}^8 & \{4,5\}^9 & \{4,6\}^{10} \\ \end{tabular} \end{center}](https://tex.z-dn.net/?f=%5Cbegin%7Bcenter%7D%0A%5Cbegin%7Btabular%7D%20%7B%7C%20c%20%7C%7C%20c%20%7C%20c%20%7C%20c%20%7C%20c%20%7C%20c%20%7C%20c%20%7C%7D%0A%26%201%20%26%202%20%26%203%20%26%204%20%26%205%20%26%206%20%5C%5C%20%5B1ex%5D%0A1%20%26%20%5C%7B1%2C1%5C%7D%5E2%20%26%20%5C%7B1%2C2%5C%7D%5E3%20%26%20%5C%7B1%2C3%5C%7D%5E4%20%26%20%5C%7B1%2C4%5C%7D%5E5%20%26%20%5C%7B1%2C5%5C%7D%5E6%20%26%20%5C%7B1%2C6%5C%7D%5E7%20%5C%5C%20%0A2%20%26%20%5C%7B2%2C1%5C%7D%5E3%20%26%20%5C%7B2%2C2%5C%7D%5E4%20%26%20%5C%7B2%2C3%5C%7D%5E5%20%26%20%5C%7B2%2C4%5C%7D%5E6%20%26%20%5C%7B2%2C5%5C%7D%5E7%20%26%20%5C%7B2%2C6%5C%7D%5E8%20%5C%5C%20%0A3%26%20%5C%7B3%2C1%5C%7D%5E4%20%26%20%5C%7B3%2C2%5C%7D%5E5%20%26%20%5C%7B3%2C3%5C%7D%5E6%20%26%20%5C%7B3%2C4%5C%7D%5E7%20%26%20%5C%7B3%2C5%5C%7D%5E8%20%26%20%5C%7B3%2C6%5C%7D%5E9%20%5C%5C%20%0A4%20%26%20%5C%7B4%2C1%5C%7D%5E5%20%26%20%5C%7B4%2C2%5C%7D%5E6%20%26%20%5C%7B4%2C3%5C%7D%5E7%20%26%20%5C%7B4%2C4%5C%7D%5E8%20%26%20%5C%7B4%2C5%5C%7D%5E9%20%26%20%5C%7B4%2C6%5C%7D%5E%7B10%7D%20%5C%5C%20%0A%5Cend%7Btabular%7D%0A%5Cend%7Bcenter%7D)
From the table it can be seen that the sums: 2 and 12 appeared only once and hence will represent the shortest bars if the distribution is represented in a bar chart.
Therefore, the <span>two sums that are represented by the shortest bars on a bar graph of this distribution</span> are 2 and 12.
The same goes for the second number cube.
The table below presents the possible outcomes of rolling two number cubes with the sum written as exponent.
From the table it can be seen that the sums: 2 and 12 appeared only once and hence will represent the shortest bars if the distribution is represented in a bar chart.
Therefore, the <span>two sums that are represented by the shortest bars on a bar graph of this distribution</span> are 2 and 12.