Answer: 2/1
Step-by-step explanation:
2/1 because you use rise/run which is moving up 2 units from (0,3) to (0,5) and then moving right 1 unit to (1,5)
The distance between Car A and car B,When Car A crosses the start line is:
distance =speed car B* time
distance=(15 m/s)(3 s)=45 m
Distance traveled by car A =x, (when the car B is at the same distance from the start line)
time of car A=t
x=10 m/st ⇒ x=10t (1)
Distance traveled by car B=x
time of car B=t-3
x=15(t-3) (2)
With the equations (1) and (2) we make a system of equations:
x=10t
x=15(t-3)
We solve this system of equations:
10t=15(t-3)
10t=15t-45
-5t=-45
t=-45 / -5
t=9
t-3=9-3=6
x=10 t=10 (9)=90
Answer: The time would be 9 seconds for Car A and 6 seconds for car B and the distance would be 90 meters.
Answer:
-9x+27
Step-by-step explanation:
Step-by-step explanation:
it will take you 15 minutes
Answer:
The complete question in the graphic
A.
After synchronized exercise, it was observed that the average proximity rate has increased, which shows that the mean of the variable is positive and there is high synchronization in the groups.
.
B.
If groups with low synchronization and high effort are compared, the result of the average will be positive, so the average proximity rate will also increase after the synchronization exercise is performed.
On the contrary, if we observe the groups that have low synchronization and low effort, the mean of the variable is negative, thus the average proximity rate after performing the exercise will decrease.
(C)
considering that the total degrees of freedom is 259, it will show 260 students in the analysis.
D.
With a significance level of 5% or 0.05, the P value is less than the significance level, which leads to differentiating the mean of the scores before and after performing the combinations.
E.
With a significance level of 1% or 0.01, the p-value is higher than the significance level. With this data we can conclude that there is no difference between the results before and after the synchronized exercise