Answer:
The fifth term is 7
Step-by-step explanation:
Looking at the graph
we have the ordered pairs
(1,5),(2,5.5),(3,6),(4,6.5),(5,7)
so
Let

The common difference in this arithmetic sequence is 0.5
The value of the fifth term is a_5
therefore
The fifth term is 7
Its D. they are equidistant from the center of the circle , just took the test and that was the answer on apex
Answer:
f'(1)=150ln(1.5)
Step-by-step explanation:
I'm not sure why you would need a table since the limit definition of a derivative (from what I'm remembering) gives you the exact formula anyway... so hopefully this at least helps point you in the right direction.
My work is in the attachment but I do want to address the elephant on the blackboard real quick.
You'll see that I got to the point where I isolated the h's and just stated the limit equaled the natural log of something out of nowhere. This is because, as far as I know, the way to show that is true is through the use of limits going to infinity. And I'm assuming that you haven't even begun to talk about infinite limits yet, so I'm gonna ask you to just trust that that is true. (Also the proof is a little long and could be a question on it's own tbh. There are actually other methods to take this derivative but they involve knowing other derivatives and that kinda spoils a question of this caliber.)
Answer:
92% Confidence Interval = [8.515, 8.605]
Step-by-step explanation:
The number of samples given is 15,this less than 30, hence, we use the t score confidence Interval
= Mean ± t score × Standard deviation/√n
Mean = 8.56 ounces
Standard deviation = 0.09 ounces
n = 15
We find the degrees of freedom = n - 1
= 15 - 1 = 14
Using the T score table
T score for 92% confidence interval with degrees of freedom 14
= 1.8875
Hence:
Confidence Interval =
= 8.56 ± 1.8875 × 0.09/√15
= 8.56 ± 0.0454010035
Confidence Interval
8.56 - 0.0454010035
= 8.5145989965
≈ 8.515
8.56 + 0.0454010035
= 8.6054010035
≈ 8.605
92% Confidence Interval = [8.515, 8.605]