Answer:
The data we have is:
The acceleration is 3.2 m/s^2 for 14 seconds
Initial velocity = 5.1 m/s
initial position = 0m
Then:
A(t) = 3.2m/s^2
To have the velocity, we integrate over time, and the constant of integration will be equal to the initial velocity.
V(t) = (3.2m/s^2)*t + 5.1 m/s
To have the position equation, we integrate again over time, and now the constant of integration will be the initial position (that is zero)
P(t) = (1/2)*(3.2 m/s^2)*t^2 + 5.1m/s*t
Now, the final position refers to the position when the car stops accelerating, this is at t = 14s.
P(14s) = (1/2)*(3.2 m/s^2)*(14s)^2 + 5.1m/s*14s = 385m
So the final position is 385 meters ahead the initial position.
The first one is the last answer because you subtract 18 and when you divide you change the inequality symbol
the second one is the last answer for the same reason
the 3rd one is x<-1.25
the 4th one is x<8/15
It’s the light blue one!!!
Answer:
a) Bar chart
b) Histogram
c) Bar chart
d) Histogram
Step-by-step explanation:
a) Trash pick-up DAY for each HOUSEHOLD in Ames - This is categorical data because, we are talking about days of the week. For instance, Household 1 might have Sunday as Trash pick-up day and that could be accumulated into frequency. Hence, BAR chart is the most appropriate.
b) Patient WAIT-TIME at ISU. This is continuous (quantitative) data. And the most appropriate is HISTOGRAM.
c) Number of trips taken during a GIVEN SCHOOL YEAR by EACH ISU STUDENT. Let say we have 5 ISU STUDENTS. Student 1 had 5 trips, student 2 had 10 trips, etc.
We want to see which student has the most and least trip in that particular school year. Although is count data but the most appropriate graphical display is BAR chart.
d) TAX BRACKET of ALL Iowa RESIDENTS. This is a continuous (quantitative) data. The most appropriate graphical display is HISTOGRAM.
