Could I know how to calculate to get the answer for 79,156 and 63,956? Thank you!

The WLJ Boat Company, which is under contract to the U3 Navy, assembles troop deployment boats. As part of its rescarch program, it completes the asscmbly of the first of a new model (PT109) of deployment boats. The Navy is impressed with the PT109 It requests that W Bost submit a proposa on he cost ot producing another six p11( gs w B at reports the t llowing cost into mat on or he tir t 1109 assembled and uses a 9 % cumulative avera e time learning model as a bass, tor tore asti direct manufacturing labor-hours for the next six PT 109s (A 90% leaming curve means b--D 152004) (Click the icon to vew the cost infomation, Road the rcguiremcnts Requirement 1. Calculate predicted total costs of producing the six T109s for the Navy (WLJ Boat wil keep the first deployment boat assembled, costed at $1.339200, as a Hegin by determining the hours used to produce the six PT109s or the Navy (Round your answers to the nearest whole number) The total cumulative tirne in labor-hours for seven PT109s is PT 109s 1 63,958 hours Now calculate the predicted total costs of producing the six PT109s. (Round your answers to the nearest whole dollar.) Direct materials Direct manufacturing labor Variable manufacturing overhead Other manufacturing overhead Total costs n model tor potential customers) Data Table 79,15 hours, therefore, the total lime to produce six Direct material cost Direct manufacturing labor tirne for first boat Direct manufacluring labor rale Variable manufacturing overhoad cost Other manufacturing overhead Tooling costsa Learning curve tor manutacturing labor time per boat S204,000 15,200 labor-hours S36 per direct manufacturing labor-hour $20 per direct manufacturing labor-hour 20% of direct manufacturing labor costs 284,000 90% cumulative average timeb Tooling can be reused at no extra cost because all of its cost has been assigned to the first deploymernt boal b using the formula for a 90% learning curve, b- In 0.90 0105361 0 693147 0.152004