Answer:
95%
Step-by-step explanation:
The z-scores associated with the limits are ...
z = (x -μ)/σ
z = ({80, 108} -94)/7 = {-14, 14}/7 = {-2, 2}
That is, we're interested in flight times that lie within 2 standard deviations of the mean. The empirical rule tells us that 95% of the distribution lies within 2 standard deviations of the mean.
Answer:
Step-by-step explanation:
If the position function is
and we are looking for time when the height is 0, we sub in a 0 for h(t) and solve for t:
and the easiest way to do this is to factor by taking the GCF of -40t:
0 = -40t(t - 5) and by the Zero Product Property,
-40t = 0 or t - 5 = 0. Solving for t, we get
t = 0 (which is before the object is launched) and
t = 5 (which is how long it takes the object to go from the ground, up to its max height, and then back to the ground again).
Your choice is c) 5
You need to flip the shapes over the x-axis and move over -8 to 0
Answer:
I believe 7 is the GTF (greatest common factor)
Answer:
C ≤ 2
Step-by-step explanation:
Given that :
Clayton is at most 2 meters above sea level ;
Let Clayton = C
Clayton's distance can be represented by the inequality :
2 meters above sea level is positive, +2
At most 2 meters above sea level means, Clayton could be anywhere between + 2 metwrs but not more than 2 meters
Hence, Clayton's distance can be represented by the inequality :
C ≤ 2