Question

What will be the output of the following códe?

x = True
y = False
z = False

if not x or y:
print (1)
elif not x or not y and z:
print (2)
elif not x or y or not x and y:
print (3)
else:
print (4) ​

Answer 1

Answer:

= True

y = False

z= False

If not x or y:

print 1

elif not x or not y and z: print 2

elif not x or y or not y and x print 3

else:

print 4

Corrected CØde:

X = True y = False

z = False

if not x or y:

print (1) elif not x or not y and z:

print (2) elif not x or y or not y and x:

print (3)

else:

print (4)

Output:

. 3

Explanation:

It's given that, 1. x = True 2. y = False

3. z = False

not x or y = not True or False

= not True True or False is true as one condition is true.]

= False [not True means False]

Therefore, the result is false. >> If block will not execute.

Now, check the second condition,

not x or not y and z

= not True or not False and False

= False or True and False

Order of evaluation is from left to right.

= (False or True) and False

= True and False

= False (As one condition is False = ]

Therefore, the result is false. >> First else if block will not execute.

Now, check the next condition,

not x or y or not y and x

= not True or False or not False and True

= False or False or True and True [not True = False and not False = True

= (False or False) or True and True

= (False or True) and True False or False is False as whole condition is False]

= True and True (True as one condition is True

= True

Therefore, result is True, >> This block will execute.

.. Output: 3

Explanation:

pls mark brainliest

Answer 2

Explanation:

It is given:

Charcoal has the initial activity, which is denoted as A_{0}=15.3A

0

=15.3 disintegrations per minute per gram.

Charcoal has the half-life, T 12=5730T12=5730 years

After a few years, the charcoal’s final activity, A = 12.3 disintegrations per minute per gram

Constant of disintegration,

\lambda=0.693 \mathrm{T} 12=0.6935370 \mathrm{y}-1λ=0.693T12=0.6935370y−1

For the action to attain 12.3 disintegrations per minute per gram, let the time taken by the sample at a time of t year.

Sample’s activity,

A=A O e-\lambda tA=AOe−λt

A=A O e-0.6935730 \times tA=AOe−0.6935730×t

\Rightarrow \mathrm{t}=1804.3 \text { years }⇒t=1804.3 years