(a) translate the argument into symbolic form and (b) determine if the argument is valid or invalid. You may compare the argument to a standard form or use a truth table.

If Nicholas Thompson teaches this course, then I will get a passing grade.

I did not get a passing grade.

∴ Nicholas Thompson did not teach the course

a landing airplane, moving in a straight line , passes over two points (p and q) which are 300 feet apart. when the airplane is above p, as seen from Q, its angle of elevation is 60 degress, and when above Q, as seen from p, its angle of elevation is 40 degress. at what distance from q will the airplane land?

Let's translate the argument into symbolic form:

P: Nicholas Thompson teaches this course.
Q: I will get a passing grade.

The argument can be represented as:

If P, then Q.
~Q.
∴ ~P.

To determine if the argument is valid or invalid, we can use a truth table:

P | Q | If P, then Q | ~Q | ~P
--------------------------------
T | T | T | F | F
T | F | F | T | F
F | T | T | F | T
F | F | T | T | T

From the truth table, we can observe that whenever the premises are true (P is false and Q is true), the conclusion is also true ( ~P is true). Therefore, the argument is valid.

To translate the given argument into symbolic form, we can use variables to represent the statements involved. Let's assign the following variables:

A: Nicholas Thompson teaches this course.
B: I will get a passing grade.

Now we can rewrite the argument in symbolic form:

If A, then B.
Not B.
∴ Not A.

To determine if the argument is valid or invalid, we can use the law of inference called modus tollens. Modus tollens states that if we have a conditional statement "If A, then B," and we know "Not B" is true, we can infer "Not A." So, in this case, the argument follows the modus tollens form.

Therefore, the argument is valid. We don't need to compare it to a standard form or use a truth table in this case.

Let p = Thompson teaches this course.

Let q = I will get a passing grade.

Then the symbolic form should be ~q --> ~p. Completely valid!

Here's why...
p......q......~q......~p......~q --> ~p
T......T.......F........F..............T
T......F.......T........F..............F
F......T.......F........T..............T
F......F.......T........T..............T <---

Notice it is logically equivalent to p --> q (i.e., the truth table of ~q --> ~p is the same as the truth table of p --> q).