Consider the following instance variable and method: private int[] nums;
/** print the elements that are in the odd index locations off an array: EX: 1, 3, 5, etc. */
public void printOddIndices()
{
/* missing code */
}

Which of the following replacements for /* missing code */ correctly implements the method printOddIndices()?

a)
for (int x : nums)
{
if (x % 2 == 1)
{
System. out. println(x);
}
}

b)
for (int x : nums)
{
if (nums[x] % 2 == 1)
{
System. out. println(x);
}
}

c)
for (int k = 0; k < nums. length; k++)
{
if (k % 2 == 0)
{
System. out. println(nums[k]);
}
}

d)
for (int k = 0; k < nums. length; k++)
{
if (nums[k] % 2 == 1)
{
System. out. println(k);
}
}

e)
for (int k = 0; k < nums. length; k++)
{
if (k % 2 == 1)
{
System. out. println(nums[k]);
}
}

Peripherals are part of the main computer. true false

Given int variables k and total that have already been declared, use a while loop to compute the sum of the squares of the first 50 counting numbers, and store this value in total. thus your code should put 11 + 22 + 33 + + 4949 + 50*50 into total. use no variables other than k and total.

Given a link with a maximum transmission rate of 32.8 mbps. only two computers, x and y, wish to transmit starting at time t = 0 seconds. computer x sends filex (4 mib) and computer y sends filey (244 kib), both starting at time t = 0. statistical multiplexing is used, with details as follows packet payload size = 1000 bytes packet header size = 24 bytes (overhead) ignore processing and queueing delays assume partial packets (packets consisting of less than 1000 bytes of data) are padded so that they are the same size as full packets. assume continuous alternating-packet transmission. computer x gets the transmission medium first. at what time (t = ? ) would filey finish transmitting? give answer in milliseconds, without units, and round to one decimal places (e.g. for an answer of 0.013777 seconds you would enter "13.8" without the quotes)