We are given this explicit formula :
In order to find eighth term let us plug n as 8 in this formula:
subtracting 8-1 first using pemdas rule
multiplying 5 and 7
subtracting 35-6
c(8) = 29
So eighth term is 29.
Answer:yes. It i s correct
Step-by-step explanation:
Problems of this sort are frequently found in physics. If you study calculus or physics you'll learn how to create the equation representing the velocity of an object in flight.
Here, you don't need to calculate velocity, but rather time. Start with this equation:
v = v0 + a t^2, where v is the velocity at time t, v0 is the initial velocity, a is the acceleration due to gravity (denoted by g instead of a), and t is the elapsed time.
You are told that v0 is 15 ft/sec. Set v = to 0, as the ball stops moving for the tiniest instant at the top of its trajectory. Use g = - 32 ft (per second squared).
Then 0 = 15 ft/sec - 32 [ft/(seconds squared)] t.
Solve this for t. This is the time required for the ball to come to a complete stop at the top of its trajectory.
Finally, multiply this time by 2, since the ball begins to fall and returns to its original height.
The answer is Angle.
Rays that share and endpoint form an angle. :)
That outlier will appear as a single blip of data, but it will be far
away from the central part of the histogram. It will not be that visible
if there is a lot of data, since 1 outlying data point will just show a
very low height compared to the central data near the mean/median/mode,
where there will be higher frequencies and higher bars of data.
If these are the choices:
<span>A.The outlier will appear as a tall bar near one side of the distribution.
B.The outlier will appear as a bar far from all of the other bars with a height that corresponds to a frequency of 1.
C.Since a histogram shows frequencies, not individual data values, the
outlier will not appear. Instead, the outlier increases the frequency
for its class by 1.
D.The outlier will appear as the tallest bar near the center of the distribution.
</span>
Then the most accurate answer will be B, since it's just a single value with a bar corresponding to a frequency of 1.