Question

For a three-year select-and-ultimate table, you are given: - q [x]​ =(1−2k)q x​ ,q [x]+1​ =(1−k)q x+1​ and q[x]+2​ =0.9qx+2​ . - l [32]​ =87.10. - l 32​ =100,l33​ =90,l 34​ =72 and l 35​ =54. Calculatel [32]+1 ​ .

Answer 1

The value of l[32]+1 is 88.10.

Step-by-step explanation:

To calculate l[32]+1, we can start by using the given equation for q[x]:

q[x] = (1 - 2k)qx

We know that l[32] = 87.10, so we can substitute this value into the equation above:

87.10 = (1 - 2k)qx

Now, we can solve for qx by dividing both sides by (1 - 2k):

qx = 87.10 / (1 - 2k)

Next, we can use the equation for q[x]+1:

q[x]+1 = (1 - k)qx + 1

Substituting qx = 87.10 / (1 - 2k), we get:

q[x]+1 = (1 - k) (87.10 / (1 - 2k)) + 1

Simplifying, we get:

q[x]+1 = 87.10 / (1 - k) + 1

Now, we can substitute the value of l[32] + 1:

l[32]+1 = 87.10 / (1 - k) + 1

Simplifying, we get:

l[32]+1 = 88.10