Question

NO LINKS!! Please help me with this problem. Part 10ff​

Answer 1

`\begin{array}c\cline{1-2} n& P\\\cline{1-2} 1& 0.5\\\cline{1-2} 2& 0.74\\\cline{1-2}3& 0.82\\\cline{1-2} 4 & 0.86\\\cline{1-2} 5& 0.89\\\cline{1-2} 6& 0.91\\\cline{1-2} 7& 0.92\\\cline{1-2} 8& 0.93\\\cline{1-2} 9& 0.94\\\cline{1-2} 10& 0.95\\\cline{1-2} \end{array}`

Explanation:

Given function:

`P=(0.5+0.9(n-1))/(1+0.9(n-1))`

To complete the table, substitute the given values of n into the given function.

`\boxed{\begin{aligned}n=1 \implies P & =(0.5+0.9(1-1))/(1+0.9(1-1))\\\\P & =(0.5+0.9(0))/(1+0.9(0))\\\\P & =(0.5+0)/(1+0)\\\\P & =(0.5)/(1)\\\\P&=0.5\end{aligned}}`

`\boxed{\begin{aligned}n=2 \implies P & =(0.5+0.9(2-1))/(1+0.9(2-1))\\\\P & =(0.5+0.9(1))/(1+0.9(1))\\\\P & =(0.5+0.9)/(1+0.9)\\\\P & =(1.4)/(1.9)\\\\P&=0.74 \; \sf (2 \; d.p.)\end{aligned}}`

`\boxed{\begin{aligned}n=3 \implies P & =(0.5+0.9(3-1))/(1+0.9(3-1))\\\\P & =(0.5+0.9(2))/(1+0.9(2))\\\\P & =(0.5+1.8)/(1+1.8)\\\\P & =(2.3)/(2.8)\\\\P&= 0.82 \; \sf (2 \; d.p.)\end{aligned}}`

`\boxed{\begin{aligned}n=4 \implies P & =(0.5+0.9(4-1))/(1+0.9(4-1))\\\\P & =(0.5+0.9(3))/(1+0.9(3))\\\\P & =(0.5+2.7)/(1+2.7)\\\\P & =(3.2)/(3.7)\\\\P&= 0.86 \; \sf (2 \; d.p.)\end{aligned}}`

`\boxed{\begin{aligned}n=5 \implies P & =(0.5+0.9(5-1))/(1+0.9(5-1))\\\\P & =(0.5+0.9(4))/(1+0.9(4))\\\\P & =(0.5+3.6)/(1+3.6)\\\\P & =(4.1)/(4.6)\\\\P&= 0.89\; \sf (2 \; d.p.)\end{aligned}}`

`\boxed{\begin{aligned}n=6 \implies P & =(0.5+0.9(6-1))/(1+0.9(6-1))\\\\P & =(0.5+0.9(5))/(1+0.9(5))\\\\P & =(0.5+4.5)/(1+4.5)\\\\P & =(5)/(5.5)\\\\P&= 0.91 \; \sf (2 \; d.p.)\end{aligned}}`

`\boxed{\begin{aligned}n=7 \implies P & =(0.5+0.9(7-1))/(1+0.9(7-1))\\\\P & =(0.5+0.9(6))/(1+0.9(6))\\\\P & =(0.5+5.4)/(1+5.4)\\\\P & =(5.9)/(6.4)\\\\P&=0.92 \; \sf (2 \; d.p.)\end{aligned}}`

`\boxed{\begin{aligned}n=8 \implies P & =(0.5+0.9(8-1))/(1+0.9(8-1))\\\\P & =(0.5+0.9(7))/(1+0.9(7))\\\\P & =(0.5+6.3)/(1+6.3)\\\\P & =(6.8)/(7.3)\\\\P&=0.93\; \sf (2 \; d.p.)\end{aligned}}`

`\boxed{\begin{aligned}n=9 \implies P & =(0.5+0.9(9-1))/(1+0.9(9-1))\\\\P & =(0.5+0.9(8))/(1+0.9(8))\\\\P & =(0.5+7.2)/(1+7.2)\\\\P & =(7.7)/(8.2)\\\\P&=0.94 \; \sf (2 \; d.p.)\end{aligned}}`

`\boxed{\begin{aligned}n=10 \implies P & =(0.5+0.9(10-1))/(1+0.9(10-1))\\\\P & =(0.5+0.9(9))/(1+0.9(9))\\\\P & =(0.5+8.1)/(1+8.1)\\\\P & =(8.6)/(9.1)\\\\P&=0.95 \; \sf (2 \; d.p.)\end{aligned}}`