Correct this code so that it takes one argument, `x`, and returns "`x` is more than zero" if `x` is positive (and nonzero), and otherwise, returns "`x` is equal to or less than zero." In both cases, replace `x` with the actual value of `x`.

Source: codewars | Difficulty: EASY

Starter code: def corrections(x):
	

Solutions:
def corrections(x):
	str = '{0} is more than zero.' if x > 0 else '{0} is equal to or less than zero.'
	return str.format(x)


def corrections(x):
	if x > 0:
		return f'{x} is more than zero.'
	else:
		return f'{x} is equal to or less than zero.'


corrections = lambda x: '%s is %s than zero.' % (x, ('equal to or less', 'more')[x > 0])


def corrections(x):
	return '%d is %s than zero.' % (x, ['equal to or less', 'more'][x > 0])


def corrections(x):
	if float(x) > 0.0:
		return str(int(x)) + ' is more than zero.'
	else:
		return str(int(x)) + ' is equal to or less than zero.'