Answer:
C
Step-by-step explanation:
In this question, we are asked to calculate the interquartile range of the speed values given in the question.
Firstly, we need to order the values of the speed from the lowest to the highest. That would be;
55 58 65 66 70 72 81
Now, we determine the first quartile position which is Q1. That would be (N+1)/2 where N is the number of data points which is 7 according to this question
We have (7 +1)/4= 8/4= 2
The value at the 2nd position counting from left is 58
For the third quartile , Q3, we have 3 * (N+1)/4 = 3 * 2 = 6
The value at the 6th position counting from left is 72
mathematically, the formula for the interquartile range is Q3 - Q1 = 72 - 58 = 14
If 3/10 was chocolate
3/10=0.3
0.3 *100=30% was chocolate
The second one and the last one
Answer:
21,27,33 that's all for now
a f(a) is the function f(x) where x is replaced by a.
So we have
(3x^2 + x + 3 - (3a^2 + a + 3)) / ( x - a)
= (3x^2 - 3a^2 + x - a) / (x = a) Answer
b (3(x + h)^2 + x + h + 3 - (3x^2 + x + 3)) / h
= (3x^2 + 6xh + 3h^2 + x + h - 3x^2 - x - 3) ) / h
= (6xh + h + 3 h^2) / h
= 6x + 3h + 1 Answer
