What value is perpendicular to a slope of m = 2? *

Year 7 benefit would be equal to gain £850 fault payout.

Likely to mean proof findings to 13 weeks being equal to 0.13 shift change

as 0.26 last 3/4 year  (P) can be proved to be 0.39 if first 38.5 weeks are used and converted same as any other year for year 7.

As we can easily compare with year 7 to

Proof

39/100 - 13/100 = 0.26

is the same as 1st 38.5 weeks less 13 first weeks

Is provable to be year 7

.

Step-by-step explanation:

8900 fault payout amount

8900 - 1150 = 7750   yr 1

7750 - 1150 = 6600   yr 2

6600 - 1150 = 5450  yr 3

5450- 1150 = 4300   yr 4

4300 - 1150 = 3150   yr 5

3150 - 1150 = 2000   yr 6

2000- 1150 = 850    yr 7

850 - 850 = 0       yr 8

p) 0.26 =  3/4 (365) = 273.75 remaining days in the year

p) 0.74 = 1/4 x 365 = 91.25 first days in the year

lets look at year 8 = 1150/850 = 1.35294118

365/ 1.35294118 = 269.782608 = 270 days approx.

This is close to an 8 month period as seen in remaining days calculus.

To find exact amount of weeks that 269.78... days is we divide by 7

269.782608 / 7  = 38.5403726 weeks =  38.5 weeks.

52/ 38.5 = 1.35 our multiplier

1.35 / 0.26 = 5.19230769 our fraction of 38.5 weeks would be 1/5 + 19/100.

= 0.39  average first 38.5 weeks.

Same as 1150/ 850  = 1.35 = 0.39 average first 269.78 days

= 39/100 (P)

0.39 / 0.26 = 1.5 scale factor increase from claiming last 3/4 year to claiming first 3/4 which means claiming first 1/4 would be 0.5 + 0.26 = 0.31 first 3 months (1/4 of year)

0.31  = first 13.5 weeks

0.39 = first 38.5 weeks  .

0.26 = last 38.5 weeks.

= difference of 0.13

= 3/20 (P) Rise

if increase is 0.15 then the expected benefit of claim rises 1/0.15 =  15% RISE  which is 3/20 rise.

39/100 - 13/100 = 0.26

is the same as 1st 38.5 weeks less 13 first weeks