the answers are wrong

Machine $x$ has initial cost $-80,000$$$,$ annual operating cost $-20,000\frac{\text{\$}}{y}$ear a salvage value $of10,000$$ and life $of2$ years. Machine $Y$ has initial cost $-$

$95,000$$$,$ annual operating cost $-15,000$$ $\frac{?}{\text{y}\text{e}\text{a}\text{r}}$ a salvage value $of30,000$$ and life $of4$ years. Comparing these mutually exclusive alternatives, the

equation that will calculate the present worth $of$ machine $xat10\%$ interes$\frac{t}{y}$ear $is$:

$a.PWx=-80,000-20,000(\frac{P}{A},10\%,4)-70,000(\frac{P}{F},10\%,4)+10,000(\frac{P}{F},10\%,2)$

$b.PWx=-80,000-20,000(\frac{P}{A},10\%,2)+10,000(\frac{P}{F},10\%,2)$

$c.PWx=-80,000-20,000(\frac{P}{A},10\%,4)-70,000(\frac{P}{F},10\%,2)+10,000(\frac{P}{F},10\%,4)$

$d.PWx=-80,000-15,000(\frac{P}{A},10\%,4)+30,000(\frac{P}{F},10\%,4)=PA+PGG(arithmetic$ gradient$)=40,n=9$ years, $A=340$ and $i=8\%$ per

year?

$a.4015$

$b.2487$

$C.3596$

$d.2996$

Approximately, what $is$ the value $of(P)ifA=540,n=19$ years, and $i=18\%$ per year?

$a.2383$

$b.3446$

$c.2871$

$d.3848$