How efficient well if we are transmitting 6bits and we need 10 to transmit the 6 bits that would be 6/10 = .6 = 60% efficient.
Answer:
The "a" Option is correct.
Explanation:
The "COUNTIF" function counts every cell that, given a condition (value), suits into it. As you want to know the number of cells that contain a value of at least 50, the condition must be properly written to get the correct answer. Unless it is a cell value (e.g. B3), the condition must always be written with quotes (""). So, the options b and c are automatically discarded.
The d option appears to be correct, but it's not. If the condition is written ">50", the function will count every cell with a value above 50. But we're searching values at least (including) 50. So the correct answer is the a option.
Answer:
def find_max(num_1, num_2):
max_val = 0.0
if (num_1 > num_2): # if num1 is greater than num2,
max_val = num_1 # then num1 is the maxVal.
else: # Otherwise,
max_val = num_2 # num2 is the maxVal
return max_val
max_sum = 0.0
num_a = float(input())
num_b = float(input())
num_y = float(input())
num_z = float(input())
max_sum = find_max(num_a, num_b) + find_max(num_y, num_z)
print('max_sum is:', max_sum)
Explanation:
I added the missing part. Also, you forgot the put parentheses. I highlighted all.
To find the max_sum, you need to call the find_max twice and sum the result of these. In the first call, use the parameters num_a and num_b (This will give you greater among them). In the second call, use the parameters num_y and num_z (This will again give you greater among them)
Answer:
See explaination
Explanation:
StackExample.java
public class StackExample<T> {
private final static int DEFAULT_CAPACITY = 100;
private int top;
private T[] stack = (T[])(new Object[DEFAULT_CAPACITY]);
/**
* Returns a reference to the element at the top of this stack.
* The element is not removed from the stack.
* atreturn element on top of stack
* atthrows EmptyCollectionException if stack is empty
*/
public T peek() throws EmptyCollectionException
{
if (isEmpty())
throw new EmptyCollectionException("stack");
return stack[top-1];
}
/**
* Returns true if this stack is empty and false otherwise.
* atreturn true if this stack is empty
*/
public boolean isEmpty()
{
return top < 0;
}
}
//please replace "at" with the at symbol
Note:
peek() method will always pick the first element from stack. While calling peek() method when stack is empty then it will throw stack underflow error. Since peek() method will always look for first element ffrom stack there is no chance for overflow of stack. So overflow error checking is not required. In above program we handled underflow error in peek() method by checking whether stack is an empty or not.
