Yes 0.007*10= 0.07
Youre correct
Answer:
The values of x for which the model is 0 ≤ x ≤ 3
Step-by-step explanation:
The given function for the volume of the shipping box is given as follows;
V = 2·x³ - 19·x² + 39·x
The function will make sense when V ≥ 0, which is given as follows
When V = 0, x = 0
Which gives;
0 = 2·x³ - 19·x² + 39·x
0 = 2·x² - 19·x + 39
0 = x² - 9.5·x + 19.5
From an hint obtained by plotting the function, we have;
0 = (x - 3)·(x - 6.5)
We check for the local maximum as follows;
dV/dx = d(2·x³ - 19·x² + 39·x)/dx = 0
6·x² - 38·x + 39 = 0
x² - 19/3·x + 6.5 = 0
x = (19/3 ±√((19/3)² - 4 × 1 × 6.5))/2
∴ x = 1.288, or 5.045
At x = 1.288, we have;
V = 2·1.288³ - 19·1.288² + 39·1.288 ≈ 22.99
V ≈ 22.99 in.³
When x = 5.045, we have;
V = 2·5.045³ - 19·5.045² + 39·5.045≈ -30.023
Therefore;
V > 0 for 0 < x < 3 and V < 0 for 3 < x < 6.5
The values of x for which the model makes sense and V ≥ 0 is 0 ≤ x ≤ 3.
C doesn’t have a value , x equals 0 , x=0
Answer:
For number 1: y=2 and x=3
Find the interquartile range for the data {5, 7, 9, 5, 6, 6, 6, 11, 11, 3, 3}
RoseWind [281]
Answer:
4
Step-by-step explanation:
i dont really know how to explain i used an algebra calculator
