Please solve this question correctly. I provided a similar question with how to solve it. please solve my version and just provide the final answer.

(Raind bo the nearest wtole nimber as niecded) The Bitmore Garage has lights in places that are difficult to reach. Management estimates that it costs about $3 to change a bulb. Standard 100 -watt bulbs with an expected life of 1000 hours are now used Standard bulbs cost 51 . A kng-life bulb that requires 90 watts for the same etfective level of light is available Long life bulbs cost $3. The bulbs that ale difficult to reach are in use for about 500 hours a month Electricity costs $0. 07/k ilowati-hour payable at the end of each month Biltmore useis a 12 percent MARR (1 percent par month) for projects involving supplies a. What minimum life for the long life bubb wouid make it the lower cost option? b. If the cost of changing bulbs is ignored, what is the minimum life for the long-life builb to be the lower cost alternative? c. If the solutions are obtained by linear interpolation of the capital rocovery factor, will the approximations understate or overstate the required life? Click the icon to view the lable of compound interest factors for disctele compounding pariods when i =1% a. Detertine the equation to find the arnual worth of the long life bulb AW=5(($0.07+90500)/(1000))+($3+$3)(A/P,1%,N) Defermine the equation to find the annual worth of the standard bulb AW=(($007+100500)/(1000))+($3+51)(A/P,18,(1000/500)))7 To find the minmum life for the long-life buits that would make it a lower cost option, sed the equations oquat to each other to solve for (AvP. 1\%, N) Simpify If necessary, identify tive compound interest tactors Aithough the value of the compound inferest facior could be found either by hand pr by using a table, for the putposes of this solution, use the given tablo to find the value Silnplify ($007100500)+1000+($3+$1)(AP,1%,1000500)$350+$4(0.50751)0.39007=($0.0790500)+1000+($3+$3)(AP,1%,N)=$315+$6(AP,1%,N)=(AP,1%,N) (Type an integer or docimal rounded to five decimal placos as needed) Use linear interpolatoon to solve for N By looking at the table, the result will be between N=2 and N=3. Determine those valus. (AP,1%,2)=0.50751 (AP,14,3)=034002) (Round to tre decimal places as needed) Use tnear interpolabon to find the value of N N==2+(32)(0.507510.39667)(0.507510.34002).2662(Roundtothreedecmalplacesasneeded) The minimum Me that would make the long-life bulb the lower cost option is hours: (Round to the nearest whole number as needed.) b. To find the minimum life that would make the long lifo bulb the kwer cost opton, ignoring the cost of changing bulbs, is a similar sel up to the prior step Simplity ($0.07100+500)1000+($1)KAV,10,1000+500)$3.50+50.507510.28584==($0.0790500)1000+($3)(AvP,1%0,N)=$3.15+$3(AP,1%,N)(AVP,1%,N) (Tree an integer ardecimal rounded to five decimal places as needed) The minimum life that would make the long. life buib the tower cost option is hours (Round to the nearest whole number as noeded) b. To find the minimum life that would make the long.life bulb the fower cost option, ignoring the cost of changing bulbs; is a similar set up to ihe prior step ($0.07100500)1000+($1)(AP,1%,1000+500)=($0.0790500)+1000+($3)(AP,1%,N)$350+$0.50751=$3.15+$3(AP,1%,N) 028584=(AP,196,N) (Type an ioteger or decimal rounded to fve decimal places as rieeded) Use linear interpolation to solve fos N By looking at the table, the resuit wil be between N=3 and N=4 Determine those values (AP,1%,3)=0.3400? (AP,1%,4)=025628 (Round to five docimal placos as neoded.) Use linear interpolation to tind the value of N N==3+(43)(0.34002028584)(0340020.25628)3.647(Roundtothreedecimalplacesasrheded) Ignoring the cost of changing bulbs, the minimum life that would make the long-afe bulb the lower cost option is (Round to the nearest whole number as neoded.) c. As the solutions are obtained by linear interpolation of the capital recovery factor, the approximations will. the required life (Raind bo the nearest wtole nimber as niecded) The Bitmore Garage has lights in places that are difficult to reach. Management estimates that it costs about $3 to change a bulb. Standard 100 -watt bulbs with an expected life of 1000 hours are now used Standard bulbs cost 51 . A kng-life bulb that requires 90 watts for the same etfective level of light is available Long life bulbs cost $3. The bulbs that ale difficult to reach are in use for about 500 hours a month Electricity costs $0. 07/k ilowati-hour payable at the end of each month Biltmore useis a 12 percent MARR (1 percent par month) for projects involving supplies a. What minimum life for the long life bubb wouid make it the lower cost option? b. If the cost of changing bulbs is ignored, what is the minimum life for the long-life builb to be the lower cost alternative? c. If the solutions are obtained by linear interpolation of the capital rocovery factor, will the approximations understate or overstate the required life? Click the icon to view the lable of compound interest factors for disctele compounding pariods when i =1% a. Detertine the equation to find the arnual worth of the long life bulb AW=5(($0.07+90500)/(1000))+($3+$3)(A/P,1%,N) Defermine the equation to find the annual worth of the standard bulb AW=(($007+100500)/(1000))+($3+51)(A/P,18,(1000/500)))7 To find the minmum life for the long-life buits that would make it a lower cost option, sed the equations oquat to each other to solve for (AvP. 1\%, N) Simpify If necessary, identify tive compound interest tactors Aithough the value of the compound inferest facior could be found either by hand pr by using a table, for the putposes of this solution, use the given tablo to find the value Silnplify ($007100500)+1000+($3+$1)(AP,1%,1000500)$350+$4(0.50751)0.39007=($0.0790500)+1000+($3+$3)(AP,1%,N)=$315+$6(AP,1%,N)=(AP,1%,N) (Type an integer or docimal rounded to five decimal placos as needed) Use linear interpolatoon to solve for N By looking at the table, the result will be between N=2 and N=3. Determine those valus. (AP,1%,2)=0.50751 (AP,14,3)=034002) (Round to tre decimal places as needed) Use tnear interpolabon to find the value of N N==2+(32)(0.507510.39667)(0.507510.34002).2662(Roundtothreedecmalplacesasneeded) The minimum Me that would make the long-life bulb the lower cost option is hours: (Round to the nearest whole number as needed.) b. To find the minimum life that would make the long lifo bulb the kwer cost opton, ignoring the cost of changing bulbs, is a similar sel up to the prior step Simplity ($0.07100+500)1000+($1)KAV,10,1000+500)$3.50+50.507510.28584==($0.0790500)1000+($3)(AvP,1%0,N)=$3.15+$3(AP,1%,N)(AVP,1%,N) (Tree an integer ardecimal rounded to five decimal places as needed) The minimum life that would make the long. life buib the tower cost option is hours (Round to the nearest whole number as noeded) b. To find the minimum life that would make the long.life bulb the fower cost option, ignoring the cost of changing bulbs; is a similar set up to ihe prior step ($0.07100500)1000+($1)(AP,1%,1000+500)=($0.0790500)+1000+($3)(AP,1%,N)$350+$0.50751=$3.15+$3(AP,1%,N) 028584=(AP,196,N) (Type an ioteger or decimal rounded to fve decimal places as rieeded) Use linear interpolation to solve fos N By looking at the table, the resuit wil be between N=3 and N=4 Determine those values (AP,1%,3)=0.3400? (AP,1%,4)=025628 (Round to five docimal placos as neoded.) Use linear interpolation to tind the value of N N==3+(43)(0.34002028584)(0340020.25628)3.647(Roundtothreedecimalplacesasrheded) Ignoring the cost of changing bulbs, the minimum life that would make the long-afe bulb the lower cost option is (Round to the nearest whole number as neoded.) c. As the solutions are obtained by linear interpolation of the capital recovery factor, the approximations will. the required life