The measures of spread include the range, quartiles and the interquartile range, variance and standard deviation. Let's consider each one by one.
<u>Interquartile Range: </u>
Given the Data -> First Quartile = 2, Third Quartile = 5
Interquartile Range = 5 - 2 = 3
<u>Range:</u> 8 - 1 = 7
<u>Variance: </u>
We start by determining the mean,
n = number of numbers in the set
Solving for the sum of squares is a long process, so I will skip over that portion and go right into solving for the variance.
5.3
<u>Standard Deviation</u>
We take the square root of the variance,
2.3
If you are not familiar with variance and standard deviation, just leave it.
8 litres (amount of 20% solution needed) and 7 litres for (amount of 50% solution needed)
<u>Step-by-step explanation:</u>
Let consider ‘x’ for 20% acid solution and (15 – x) for 50% acid solution. And so, the equation would be as below,
20% in x + 50% in (15 – x) = 15 litres of 34%
Convert percentage values, we get
0.20(x) + 0.50 (15 – x) = 15 (0.34)
0.20 x + 7.5 – 0.50 x = 5.1
-0.3 x + 7.5 = 5.1
0.3 x = 7.5 – 5.1
0.3 x = 2.4
Apply ‘x = 8’ value in (15 – x) we get,
15 – 8 = 7 litres
The value of 7 litres for (amount of 50% solution needed)
The answer is A(45) = -211
Answer:
It is a linear function because the degree is 1. The function is also neither odd nor even.
