Advertisements
Advertisements
प्रश्न
Using interpolation estimate the business done in 1985 from the following data
Year | 1982 | 1983 | 1984 | 1986 |
Business done (in lakhs) |
150 | 235 | 365 | 525 |
उत्तर
Here the intervals are unequal.
By Lagrange’s formula we have,
x0 = 1982
x1 = 1983
x2 = 1984
x3 = 1986
y0 = 150
y1 = 235
y2 = 365
y3 = 525 and x = 1985.
y = `"f"(x) = ((x - x_1)(x - x_2)(x - x_3))/((x_0 - x_1)(x_0 - x_2)(x_0 - x_3)) xx y_0 + ((x - x_0)(x - x_2)(x - x_3))/((x_1 - x_0)(x_1 - x_2)(x_1 - x_3)) xx y_1 + ((x - x_0)(x - x_1)(x - x_3))/((x _2 - x_0)(x_2 - x_1)(x_2 - x_3)) xx y_2 + ((x - x_0)(x - x_1)(x - x_2))/((x_3 - x_0)(x_3 - x_1)(x_3 - x_2)) xx y_3`
y = `((1985 - 1983)(1985 - 1984)(1985 - 1986))/((1982 - 1983)(1982 - 1984)(1982 - 1986)) xx 150 + ((1985 - 1982)(1985 - 1984)(1985 - 1986))/((1983 - 1982)(1983 - 1984)(1983 - 1986)) xx 235 ((198 - 982)(1985 - 1983)(985 - 1986))/((984 - 1982)(1984 - 1983)(1984 - 1986)) xx 365 + ((1985 - 1982)(1985 - 1983)(1985 - 1984))/((1986 - 1982)(1986 - 1983)(1986 - 1984)) xx 525`
= `((2)(1)(-1))/((-1)(-2)(-4)) xx 150 + ((3)(1)(-1))/((1)(-1)(-3)) xx 235 + ((3)(2)(-1))/((2)(1)(-)) xx 365 + ((3)(2)(1))/((4)(3)(2)) xx 525`
= `(-2 xx 150)/(-8) + (-3 xx 235)/3 + (-6 xx 365)/(-4) xx (6 xx 525)/24`
= 37.5 – 235 + 547.5 + 131.25
= 716.125 – 235
= 481.25
∴ Business done in the year 1985 is 481.25 lakhs.
APPEARS IN
संबंधित प्रश्न
The population of a city in a censes taken once in 10 years is given below. Estimate the population in the year 1955.
Year | 1951 | 1961 | 1971 | 1981 |
Population in lakhs |
35 | 42 | 58 | 84 |
In an examination the number of candidates who secured marks between certain intervals was as follows:
Marks | 0 - 19 | 20 - 39 | 40 - 59 | 60 - 79 | 80 - 99 |
No. of candidates |
41 | 62 | 65 | 50 | 17 |
Estimate the number of candidates whose marks are less than 70.
The following data gives the melting point of a alloy of lead and zinc where ‘t’ is the temperature in degree c and P is the percentage of lead in the alloy.
P | 40 | 50 | 60 | 70 | 80 | 90 |
T | 180 | 204 | 226 | 250 | 276 | 304 |
Find the melting point of the alloy containing 84 percent lead.
Using interpolation, find the value of f(x) when x = 15
x | 3 | 7 | 11 | 19 |
f(x) | 42 | 43 | 47 | 60 |
Choose the correct alternative:
For the given data find the value of Δ3y0 is
x | 5 | 6 | 9 | 11 |
y | 12 | 13 | 15 | 18 |
A second degree polynomial passes though the point (1, –1) (2, –1) (3, 1) (4, 5). Find the polynomial
Find the missing figures in the following table:
x | 0 | 5 | 10 | 15 | 20 | 25 |
y | 7 | 11 | - | 18 | - | 32 |
From the following data find y at x = 43 and x = 84.
x | 40 | 50 | 60 | 70 | 80 | 90 |
y | 184 | 204 | 226 | 250 | 276 | 304 |
The area A of circle of diameter ‘d’ is given for the following values
D | 80 | 85 | 90 | 95 | 100 |
A | 5026 | 5674 | 6362 | 7088 | 7854 |
Find the approximate values for the areas of circles of diameter 82 and 91 respectively
If u0 = 560, u1 = 556, u2 = 520, u4 = 385, show that u3 = 465