The answer is
the vector PQ has (4-1, 8-2) as components, so it is vecPQ= (3, 6)
using the definition
the distance PQ= sqrt ( 3²+6²)= sqrt(9+36)=sqrt(45)
To find it, evaluate it at the endpoints and the vertex
in form
f(x)=ax²+bx+c
the x value of the vertex is -b/2a
given
c(t)=1t²-10t+76
x value of vertex is -(-10)/1=10
evaluate c(0) and c(13) and c(10)
c(0)=76
c(13)=115
c(10)=76
it reached minimum in 2000 and 2010
porbably teacher wants 2010
the min value is $76
Answer:
15.6
Step-by-step explanation:
First, multiply the midpoint of each class by its frequency, as follows:
Class midpoint frequency midpoint*frequency
0-9 4.5 24 4.5*24 = 108
10-19 14.5 20 14.5*20 = 290
20-29 24.5 32 24.5*32 = 784
Total 76 1182
The mean is computed as the division between the addition of the "midpoint*frequency" column by the addition of "frequency" column.
mean = 1182/76 ≈ 15.6
Here is a graph. When you use a graphing calculator, it isn't always necessary to solve for y.
Note different colors are used for the different problems. Problem 2 has a dashed line, as described by "Remember, ...".
Answer:-2
Step-by-step explanation:
