In a situation where data are known to three significant figures, we write 6.379
ID: 2161428 • Letter: I
Question
In a situation where data are known to three significant figures, we write 6.379 m = 6.38 m and 6.374 m = 6.37 m. When a number ends in 5, we arbitrarily choose to write 6.375 m = 6.38 m. We could equally well write 6.375 m = 6.37 m, "rounding down" instead of "rounding up," because we could change the number 6.375 by equal increments in both cases. Now consider an order-of-magnitude estimate, in which we consider factors rather than increments. We write 500 m ~ 10^3 m because 500 differs from 100 by a factor of 5 while it differs from 1000 by only a factor of 2. We write 437 m ~ 10^3 m and 305 m ~ 10^2 m. What distance differs from 1000 m and from 10000 m by equal factors, so that we could equally well choose to represent its order of magnitude either as ~ 10^3 m or as ~ 10^4 m?Explanation / Answer
let that no. be x. so, x lies between 1000 and 10 000, such that 1000/x=x/10000 =>x^2=10^(3+4) =>x=10^(7/2)=3162.28