A coin is unfair if the number of head outcomes satisfies  | h − 50/5 |>=1.645 What value of h prove the coin is unfair if tossed 100x?

First, we need to correct your notation.

A coin is unfair if the number of head outcomes satisfies |(h - 50)/5| ≥ 1.645

Because it is an absolute value, we have to consider two cases.

The positive case is:

(h - 50)/5 ≥ 1.645

h - 50 ≥ 8.225

h ≥ 58.225

The negative case is:

-[(h - 50)/5] ≥ 1.645

(h - 50)/5 ≤ -1.645

h - 50 ≤ -8.225

h ≤ 50 - 8.225

h ≤ 41.775

First, you can only have an integer number of heads, so we can change these to:

h ≤ 41 or h ≥ 59

But we can't go below 0 heads or above 100 heads.

Answer:

0 ≤ h ≤ 41 or 59 ≤ h ≤ 100

In English, if you get from 0 to 41 heads (inclusive) or 59 to 100 heads (inclusive), assume the coin is unfair.

Is that supposed to be:

|(h - 50) / 5| ≥ 1.645

(note the difference as what you have simplifes to |h - 10| ≥ 1.645)

In an equation, if we remove an absolute value we do a ± to the other side, as in this example:

|x| = 2

x = ±2

In an inequality, we do the same thing but will have to flip the sign on the - side.  So that inequlity turns into the following two:

(h - 50) / 5 ≥ 1.645 and (h - 50) / 5 ≤ -1.645

Now solve both inequalities.  Start by multiplying both sides by 5:

h - 50 ≥ 8.225 and h - 50 ≤ -8.225

Now add 50 to both sides:

h ≥ 58.225 and h ≤ 41.775

Since you can't have a fraction of a coin flip, we'll round up (or down in the case of the less than) to the nearest whole values:

h ≥ 59 and h ≤ 41

So this says the coin is fair if you get anywhere between 42 and 58 heads out of the 100.  If you get outside of that range, it's considered an unfair coin.