Answer: B. Exponential. There is a constant rate of decay or decrease.
The y-values decrease by 1/4 of the number that comes before every time.
Answer:
Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795
Step-by-step explanation:
Given
See attachment for class
Solving (a): Fill the midpoint of each class.
Midpoint (M) is calculated as:
Where
Lower class interval
Upper class interval
So, we have:
Class 63-65:
Class 66 - 68:
When the computation is completed, the frequency distribution will be:
Solving (b): Mean and standard deviation using 1-VarStats
Using 1-VarStats, the solution is:
<em>See attachment for result of 1-VarStats</em>
5 because this is equivalent to 9-4 which is 5 meters, this is because he is going up 9 and then goes down 4, do 9-4, which is 5
Answer:
(1, 0) and (3, 0)
Step-by-step explanation:
y=2x²-8x+6
x- intercept when y=0
- 2x²-8x+6=0
- x²-4x+4=1
- (x-2)²=1
- x-2=1 ⇒ x= 3
- x-2= -1 ⇒ x= 1
Looking at the graph closely we can see that the heart rate increases from 0 to 6 min then became steady from 6 to 25 min then finally decreases from 25 to 30 min. Therefore the correct answer is:
“The heart rate increases for 6 minutes, remains constant for 19 minutes, and then gradually decreases for 5 minutes.”
In real life cardiac exercises, we can interpret that the period from 0 to 6 min is the period where the person is still warming up thus leading to an increase in heart rate. At 6 min, the person is fully warmed up hence reaching a stable heart rate. Then at 25 min, the person starts cooling down which means that the exercise is ending soon.
