Suppose you roll two number cubes and find the probability distribution for the sum of the numbers. Which two sums are represent
1 answer:
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.
You might be interested in
90 degrees you are looking to your side
180 degrees you are looking behind you
around origin of 0,0
the image is flipped into the negative world if it is in posiitve or vice versa
<h2>Answer:</h2>
In the 5th year
<h2>Step-by-step explanation:</h2>For the first year, the salary is 1.2million = 1,200,000
For the second year, the salary is 1.2million + 75000 = 1,200,000 + 75,000 = 1,275,000
.
.
.
For the last year, the salary is 1.5million = 1,500,000
This gives the following sequence...
1,200,000 1,275,000 . . . 1,500,000
This follows an arithmetic progression with an increment of 75,000.
<em>Remember that,</em>
The last term, L, of an arithmetic progression is given by;
L = a + (n - 1)d ---------------(i)
<em>Where;</em>
a = first term of the sequence
n = number of terms in the sequence (which is the number of years)
d = the common difference or increment of the sequence
<em>From the given sequence,</em>
a = 1,200,000 [which is the first salary]
d = 75,000 [which is the increment in salary]
L = 1,500,000 [which is the maximum salary]
<em>Substitute these values into equation (i) as follows;</em>
1,500,000 = 1,200,00 + (n - 1) 75,000
1,500,000 - 1,200,000 = 75,000(n-1)
300,000 = 75,000(n - 1)
4 = n - 1
n = 5
Therefore, in the 5th year the maximum salary will be reached.