Okay, all of these pairs add up to 1 or 100%, except for one pair. I'll convert them all to decimals, so all must add up to 1.
3/8 = 0.375
0.625 + 0.375 = 1
62% = 0.62
0.38 + 0.62 = 1
7/8 = 0.875
0.875 + 0.125 = 1
70% = 0.70
1/3 <span>≈</span> 0.33
0.70 + 0.33 = 1.03
So the last pair of probabilities(70%, 1/3) does not belong with the other three because it does not add up to 1 like the others.
Answer: |p-72% |≤ 4%
Step-by-step explanation:
Let p be the population proportion.
The absolute inequality about p using an absolute value inequality.:
, where E = margin of error,
= sample proportion
Given: A poll result of 72% with a margin of error of 4% indicates that p is most likely to be between 68% and 76% .
|p-72% |≤ 4%
⇒ 72% - 4% ≤ p ≤ 72% +4%
⇒ 68% ≤ p ≤ 76%.
i.e. p is most likely to be between 68% and 76% (.
Answer:
[?] = 4
(the complete term is
)
Step-by-step explanation:

Answer:
No
Step-by-step explanation:
To obtain the inverse function
let y = f(x) and rearrange making x the subject, that is
y = 3x + 1 ( subtract 1 from both sides )
y - 1 = 3x ( divide both sides by 3 )
= x
Change y back into terms of x, thus
(x) = 
The user corrected that the scale of a drawing of a park reads: 5 miles to 1 cm , and we know that the park measures 1,600 square meters (user insisted that this measure is given in square meters and not square miles).
So we have to convert the 1600 square meters into miles, knowing that 1 meter is the same as: 0.000621371 miles
then meters square will be equivalents to:
1 m^2 = (0.000621371 mi)^2
then 1600 m^2 = 0.00061776 mi^2
now, since 5 miles are represented by 1 cm, then 25 square miles will be represented by 1 square cm
and therefore 0.00061776 square miles will be the equivalent to:
0.00061776 / 25 cm^2 = 0.000024710 cm^2
So and incredibly small number of square cm.
I still believe that some of the information you gave me are not in meters but in miles. (For example, the park may not be in square meters but in squared miles). The park seems to have the size of a house according to the info.