Answer to david has scores of 79, 96, 63, 54 on his algebra tests. Use an inequality to find the minimum score he can make on the final exam to pass the course with an average of 70 or higher, given that the final counts as 3 tests.

Mean = ∑x/n

70 ≤ (79+96+63+54+x)/5

Solve for x.

To find the minimum score David needs on the final exam, we can use the concept of weighted averages.

Let's assume that the final exam score is denoted by "x". Since the final exam counts as three tests, we will consider it as three scores of "x" each.

Now, we can calculate David's average algebra test score including the final exam as follows:

Average = (Sum of scores) / (Total number of tests)

Since David has four algebra test scores plus the three scores from the final exam, we have a total of seven scores.

Average = (79 + 96 + 63 + 54 + x + x + x) / 7

To pass the course with an average of 70 or higher, we can set up the following inequality:

Average >= 70

Substituting the expression for the average, we get:

(79 + 96 + 63 + 54 + x + x + x) / 7 >= 70

Now, we can solve this inequality to find the minimum score David needs on the final exam. Let's simplify and solve:

(292 + 3x) / 7 >= 70

Multiply both sides by 7 to get rid of the denominator:

292 + 3x >= 490

Subtract 292 from both sides:

3x >= 198

Divide both sides by 3:

x >= 66

Therefore, David needs to score a minimum of 66 on the final exam to pass the course with an average of 70 or higher.