Answer:


The confidence interval of standard deviation is:
to 
Step-by-step explanation:
Given

See attachment for the formatted data
Solving (a): The mean
This is calculated as:

So, we have:




Solving (b): The standard deviation
This is calculated as:




--- approximated
Solving (c): 95% confidence interval of standard deviation
We have:

So:



Calculate the degree of freedom (df)



Determine the critical value at row
and columns
and 
So, we have:
---- at 
--- at 
So, the confidence interval of the standard deviation is:
to 
to 
to 
to 
So it is asking you to group like term so
x terms can be grouped/added/subtracted to other x terms, but not to x^2 or x^3 terms
x^2 terms to x^2 and so on so
1. 9-3k+5k=
9+(5k-3k)=
9+2k
2. k^2+2k+4k=
k^2+(2k+4k)=
k^2+6k=
Answer:
1a. 495 miles. 1b. 184.8 miles. 2a. 68 km. 2b. 32 kg
Step-by-step explanation:
<h3>
Answer:</h3>
- C. (9x -1)(x +4) = 9x² +35x -4
- B. 480
- A. P(t) = 4(1.019)^t
Step-by-step explanation:
1. See the attachment for the filled-in diagram. Adding the contents of the figure gives the sum at the bottom, matching selection C.
2. If we let "d" represent the length of the second volyage, then the total length of the two voyages is ...
... (d+43) + d = 1003
... 2d = 960 . . . . . . . subtract 43
... d = 480 . . . . . . . . divide by 2
The second voyage lasted 480 days.
3. 1.9% - 1.9/100 = 0.019. Adding this fraction to the original means the original is multiplied by 1 +0.019 = 1.019. Doing this multiplication each year for t years means the multiplier is (1.019)^t.
Since the starting value (in 1975) is 4 (billion), the population t years after that is ...
... P(t) = 4(1.019)^t