Question 1192723
To solve this, we'll use the concept of the complement of the binomial distribution.

* **Probability of not seeing a hawk on a single day:** 1 - 0.35 = 0.65

* **Probability of not seeing a hawk on 'n' consecutive days:** 0.65^n

* **Probability of seeing at least one hawk in 'n' days:** 1 - 0.65^n

We want to find the minimum 'n' such that:

1 - 0.65^n >= 0.968

0.65^n <= 0.032

Taking the natural logarithm of both sides:

n * ln(0.65) <= ln(0.032)

n >= ln(0.032) / ln(0.65) 

n >= 7.85

Since 'n' must be an integer, the minimum number of days a person must watch is **8 days**.