Question 1209734
This is a binomial probability problem. Here's how to solve it:

* **n** (number of trials) = 50
* **p** (probability of success - spayed/neutered) = 0.84
* **q** (probability of failure - not spayed/neutered) = 1 - p = 0.16

The binomial probability formula is:  P(x) = (nCx) * p^x * q^(n-x)

Where nCx represents "n choose x" (the binomial coefficient).

**(a) Exactly 44 are spayed/neutered:**

P(x = 44) = (50C44) * (0.84)^44 * (0.16)^6
P(x = 44) = 1,478,745,000 * 0.002011 * 0.00001678
P(x = 44) ≈ 0.0049

**(b) At most 43 are spayed/neutered:**

This means 0 to 43 are spayed/neutered.  It's a cumulative probability.  We can use a binomial cumulative distribution function (CDF) calculator or statistical software for this.  It's the sum of probabilities from x=0 to x=43.

P(x ≤ 43) ≈ 0.8878 (using a calculator or software)

**(c) At least 43 are spayed/neutered:**

This means 43 to 50 are spayed/neutered.  We can use the complement rule:

P(x ≥ 43) = 1 - P(x < 43) = 1 - P(x ≤ 42)

Use a binomial CDF calculator:

P(x ≥ 43) = 1 - 0.8284
P(x ≥ 43) ≈ 0.1716

**(d) Between 37 and 41 (inclusive):**

This means 37, 38, 39, 40, and 41 are spayed/neutered.  We can use the CDF:

P(37 ≤ x ≤ 41) = P(x ≤ 41) - P(x ≤ 36)

Use a binomial CDF calculator:
P(37 ≤ x ≤ 41) = 0.4072 - 0.0150
P(37 ≤ x ≤ 41) ≈ 0.3922

**Summary of Answers:**

* (a) P(x = 44) ≈ 0.0049
* (b) P(x ≤ 43) ≈ 0.8878
* (c) P(x ≥ 43) ≈ 0.1716
* (d) P(37 ≤ x ≤ 41) ≈ 0.3922