This is an incomplete question. The complete question is given below:
Mike has never used slide presentation software before but he needs to create a presentation by the end of the week what resource would be most helpful to mike
a. The 350-page printed manual from the slide presentation software publisher
b. A free tutorial the slide presentation software publisher has posted on the company website
c. A trouble-shooting website created by a third party
d. The 350-page online manual from the slide presentation software publisher
Answer:
b - A free tutorial the slide presentation software publisher has posted on the company website
Explanation:
As Mike has a short time and no prior experience with a slide software, then in this scenario, the best, simplest and fastest way to learn and create a presentation a free tutorial which the slide presentation software publisher has posted on the company website as this is the same company that has created this particular software so he can be rest-assured that the resource he is relying on is authentic and up-to-date with information on latest features.
Moreover, it's efficient and quick way to learn from a free tutorial rather than from 350-page printed or online manual especially for a beginner.
Besides, his purpose is to create the presentation using the software and not trouble-shooting so trouble-shooting website created by a third party is not useful for him and it also might not be authentic or updated as well.
Solution :
public class NewMain {
public_static void main_(String[] args) {
boolean[] _locker = new boolean_[101];
// to set all the locks to a false NOTE: first locker is the number 0. Last locker is the 99.
for (int_i=1;i<locker_length; i++)
{
locker[i] = false;
}
// first student opens all lockers.
for (int i=1;i<locker.length; i++) {
locker[i] = true;
}
for(int S=2; S<locker.length; S++)
{
for(int k=S; k<locker.length; k=k+S)
{
if(locker[k]==false) locker[k] = true;
else locker[k] = false;
}
}
for(int S=1; S<locker.length; S++)
{
if (locker[S] == true) {
System.out.println("Locker " + S + " Open");
}
/* else {
System.out.println("Locker " + S + " Close");
} */
}
}
}
Answer:



So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %
Explanation:
Previous concepts
Input/output operations per second (IOPS, pronounced eye-ops) "is an input/output performance measurement used to characterize computer storage devices like hard disk drives (HDD)"
Solution to the problem
For this case since we have 4GB, but 512 MB are destinated to the operating system, we can begin finding the available RAM like this:
Available = 4096 MB - 512 MB = 3584 MB
Now we can find the maximum simultaneous process than can use with this:

And then we can find the maximum wait I/O that can be tolerated with the following formula:

The expeonent for p = 14 since we got 14 simultaneous processes, and the rate for this case would be 99% or 0.99, if we solve for p we got:



So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %
For example when there is research on how many people shoplift in a shopping centre per week. These numbers could show a increase in the number of shoplifters per week.
Answer:
Step 1 : Create an Indicator Variable for metro cities using formula mentioned in formula bar.
Step 2: Filter the Data on Metro cities i.e. select only those cities with Metro Indicator 1.
Step 3: Paste this filtered data to a new sheet.
Step 4: Go to Data - Data Analysis - Regression
Step 5: Enter the range of Y-variable and X-variable as shown. Select Output range and click on residuals. It will give you Output Summary and the Predicted Values along with Residuals
Please see attachment