Answer:
25% chance.
Explanation:
There is 25% chance it'll be h*m*zygous red, 50% chance it'll be h*t*rozygous red, and 25% chance it'll be h*m*zygous wh*te.
(Censored because it wouldn't let me post it for some reason. Honestly hope this doesn't work.)
Answer:
public static void removeInRange(List<Integer> list, int value, int start, int end) {
for (int i = end - 1; i >= start; i--) {
if (list.get(i) == value) {
list.remove(i);
}
}
System.out.println(list);
}
Explanation:
- Create a method named <em>removeInRange</em> that takes four parameters, a list, an integer number, a starting index and an ending index
- Inside the method, initialize a <u>for loop</u> that iterates between starting index and ending index
- If any number between these ranges is equal to the given <em>value</em>, then remove that value from the list, using <u>remove</u> method
- When the loop is done, print the new list
Answer:
Option a. int max = aList.get(0); for (int count = 1; count < aList.size(); count++) { if (aList.get(count) > max) { max = aList.get(count); } }
is the correct code snippet.
Explanation:
Following is given the explanation for the code snippet to find largest value in an integer array list aList.
- From the array list aList, very first element having index 0 will be stored in the variable max (having data type int).
- By using for starting from count =1 to count = size of array (aList), we will compare each element of the array with first element of the array.
- If any of the checked element get greater from the first element, it gets replaced in the variable max and the count is increased by 1 so that the next element may be checked.
- When the loop will end, the variable max will have the greatest value from the array aList.
i hope it will help you!
Hope this helps you.. Sorry for the handwriting.
