Answer:
0.9451
Step-by-step explanation:
Remaining question? <em>"How is the result affected by the additional information that the survey subjects volunteered torespond?"</em>
Probability that at least 1 user is more careful about personal information when using a public Wi-Fi hot spot is:
P(X≥1) = 1 − P(X<1)
= 1 − P(X=0)
= 1 - [(3,0) (0.62)^0 (1-0.62)^3-0
= 1 - 0.054872
= 0.945128
= 0.9451
Thus, the probability that among three randomly selected Internet users; at least one is more careful about personal information when using a public Wi-Fi hotspot is 0.9451
Turn it into slope-intercept form.
2y = -x + 14
y = -x/2 + 7
The slope is -1/2.
Perpendicular lines have slopes that are opposite reciprocal, so the slope of the perpendicular line is 2.
<span>let P(n)=4n² / n²+10000n
when n ----------> infinity, the main rule is as follow
lim </span><span>P(n)=4n² / n²+10000n = lim </span>P(n)=4n² / n²= lim 4= 4
n ----------> infinity <span> n ----------> infinity </span><span>n ----------> infinity
finally
</span>
<span>Lim->infinity 4n^2/n^2+10000n
= 4</span>
Answer:
26 bags
Step-by-step explanation:
A bag of cat chows = 30 lbs
Amount of cat chows needed in a week = 15 lbs = ½ bag of cat chow
There are at least 52 weeks in a year.
The least number of cat chows bags needed at the shelter in 1 year = 52 weeks × ½ bag of chow
= 52 × ½
= 52/2
= 26 bags of cat chow
At least, 26 bags would be needed in 1 year (52 weeks) at this wild life shelter if 15 lbs is consumed weekly.
