A cost-saving microwave imaging technique may be able to identify cracks in concrete or cement-based materials before a costly structural failure occurs. If, by using this technique, a construction company saves $100,000 in year 1, $103,000 in year 2, and amounts increasing by $3,000 each year for four years, what is the

(a) equivalent present worth of the savings at an interest rate of 5% per year and

(b) equivalent annual worth of the savings at an interest rate of 5% per year?