An insurance policy charges a regular risk premium, , at Years 0,1,2,3,…,9 from today. The policy pays a claim of 2 (i. e. × ) in 15 Years from today. Calculate the value of (to 2 decimal places), assuming an interest rate of 2% per annum.

x = 12.29

Step-by-step explanation:

present value of the claim = x² / (1 + 2%)¹⁵ = 0.743x²

present value of the 10 premiums:

x + x/1.02 + x/1.02² + x/1.02³ + x/1.02⁴ + x/1.02⁵ + x/1.02⁶ + x/1.02⁷ + x/1.02⁸ + x/1.02⁹ = x + 0.9504x + 0.9612x + 0.9423x + 0.9238x + 0.9057x + 0.888x + 0.8706x + 0.8535x + 0.8368x = 9.1323x

0.743x² = 9.1323x

x² / x = 9.1323 / 0.743

x = 12.29

Answer

x = 12.29

Step-by-step explanation:

Given that the policy claims x² in 15 years, at 2% = 0.02, the present value of the claim is given as:

x²/(1 + 0.02)^15

= x²(1/1.02)^15

= 0.743x²

The present value of the 10 premiums:

x + x/1.02 + x/1.02² + x/1.02^3 + x/1.02^4 + x/1.02^5 + x/1.02^6 + x/1.02^7 + x/1.02^8 + x/1.02^9

= x + 0.9504x + 0.9612x + 0.9423x + 0.9238x + 0.9057x + 0.888x + 0.8706x + 0.8535x + 0.8368x

= 9.1323x

Now,

0.743x² = 9.1323x

x²/x = 9.1323/0.743

x = 12.29

This is what we are looking for.

l.ln.lln.k.nkl

step-by-step explanation:

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3