The $alt$ attribute is used inside the $img$ tag in HTML.

It provides alternative text for an image.

Now check each statement.

Statement A:

If a user has vision impairment, then screen readers can read the $alt$ text aloud.

So, the user can understand what the image represents.

Therefore, statement A is correct.

Statement B:

If the image does not load because of slow internet, wrong image path or broken link, then the browser can display the $alt$ text.

Therefore, statement B is correct.

Statement C:

Search engines cannot understand images like humans.

So, $alt$ text helps search engines understand the image content.

It also gives SEO benefits.

Therefore, statement C is correct.

Hence, A, B and C are correct.

Given function takes an unsigned integer $N$.

Inside the function:

$counter=0$

Now the loop runs while:

$N>0$

In each iteration:

$counter += N\ \&\ 1$

Here, $N\ \&\ 1$ checks the last bit of $N$.

If the last bit is $1$, then $counter$ increases by $1$.

If the last bit is $0$, then $counter$ does not increase.

After that:

$N=N>>1$

This shifts the bits of $N$ one position to the right.

So, the loop checks all bits of $N$ one by one.

Therefore, $counter$ stores the number of $1$ bits in the binary representation of $N$.

Finally, the function returns:

$counter==1$

This condition is true only when the binary representation of $N$ contains exactly one $1$.

A number which has exactly one $1$ in binary form is a power of $2$.

For example:

$2=(10)_2$

$4=(100)_2$

$8=(1000)_2$

So, the function returns $1$ if $N$ is a power of $2$, otherwise it returns $0$.

All attributes are single and atomic.

So, relation $R$ is in $1NF$.

Hence, statement A is true.

Now find key:

$x\to y$ and $z\to w$

So,

$(xz)^+=\{x,z,y,w\}$

Thus, $xz$ is a candidate key.

Hence, statement C is true.

For $2NF$, no non-prime attribute should depend on a proper part of candidate key.

But here,

$x\to y$

and

$z\to w$

Here $y$ and $w$ are non-prime attributes, and they depend on parts of key $xz$.

So, partial dependency exists.

Therefore, $R$ is not in $2NF$.

Hence, statement B is false.

Therefore, A and C only are correct.