Answer:
n := length(A)
repeat
swapped := false
for i := 1 to n-1 inclusive do
<em> /* if this pair is out of order */</em>
if A[i-1] > A[i] then
<em> /* swap them and remember something changed */</em>
swap(A[i-1], A[i])
swapped := true
end if ⇒
end for
until not swapped
end procedure
Explanation:
This is pseudocode
Answer:
see explaination
Explanation:
#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
double temp1,temp2,inc,cel;
int i=1;
while(i==1)
{
i=0;
cin>>temp1>>temp2>>inc;
if(temp2<temp1||inc<=0)
{
i=1;
cout<<"Starting temperature must be <= ending temperature and increment must be >0.0\n";
}
}
cout<<endl;
cout<<setw(15)<<"Fahrenheit"<<setw(15)<<"Celsius";
while(temp1<=temp2)
{
cel=(temp1-32)/1.8;
cout<<endl;
cout<<fixed<<setprecision(3)<<setw(15)<<temp1<<setw(15)<<cel;
temp1+=inc;
}
}
Please kindly check attachment for output.
Answer:
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 0.964
If the survey subjects <em>volunteered</em> to respond , then those with the strongest opinions are most likely respond. The survey sample is then not randomly selected, the survey may have a <em>response bias.</em>
Explanation:
Let P(at least one is more careful about personal information when using a public Wi-Fi hotspot) denote 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, then we have the equation
P(at least one is more careful about personal information when using a public Wi-Fi hotspot) = 1 - P(none of the selected users is more careful about personal information when using a public Wi-Fi hotspot)
P(none of the selected users is more careful about personal information when using a public Wi-Fi hotspot) = ≈ 0.036
P(at least one is more careful about personal information when using a public Wi-Fi hotspot) = 1 - 0.036 = 0.964