Questions: Each letter of the alphabet is printed on an index card. After removing the cards that spell the word "vowel," what is the theoretical probability of randomly choosing a vowel? Explain. There are 12 cats and 8 dogs in the pet store. a. What is the theoretical probability that a randomly chosen pet is a cat? b. Two days later, there are half as many cats in the pet store. The theoretical probability that a randomly chosen pet is a cat is half that of two days earlier. How many dogs are in the pet store two days later?

The theoretical probability of randomly choosing a vowel after removing the cards that spell "vowel" is: \[ \boxed{\frac{1}{7}} \]

There are 12 cats and 8 dogs, so the total number of pets is: \[ 12 + 8 = 20 \]

The probability \( P \) of randomly choosing a cat is: \[ P = \frac{\text{Number of cats}}{\text{Total number of pets}} = \frac{12}{20} = \frac{3}{5} \]

The theoretical probability that a randomly chosen pet is a cat is: \[ \boxed{\frac{3}{5}} \]

Two days later, there are half as many cats. The original number of cats was 12, so the new number of cats is: \[ \frac{12}{2} = 6 \]

The new probability \( P_{\text{new}} \) of randomly choosing a cat is half of the original probability: \[ P_{\text{new}} = \frac{1}{2} \times \frac{3}{5} = \frac{3}{10} \]

Let \( D \) be the number of dogs two days later. The total number of pets two days later is \( 6 + D \). The probability of choosing a cat is: \[ \frac{6}{6 + D} = \frac{3}{10} \]

Cross-multiply to solve for \( D \): \[ 6 \times 10 = 3 \times (6 + D) \] \[ 60 = 18 + 3D \] \[ 60 - 18 = 3D \] \[ 42 = 3D \] \[ D = 14 \]

The number of dogs in the pet store two days later is: \[ \boxed{14} \]