A distance was observed four times as 248.13, 248.06, 248.18, 248.15. The observations were given a weight of 3 1, 2, and 3, respectively, but the observer. a. Calculate the weighted mean for line BC. b. Calculate the weighted mean of BC if weights are revised as 1, 1, 3, & 3 respectively.

a. 248.14

b. 220.576

Step-by-step explanation:

The weighted mean can be calculated as

Xbarw= sumwx/sumw

where x are the observations and w are the weights of the observations.

a.

x              w         wx

248.13      3        744.39

248.06     1         248.06

248.18      2        496.36

248.15      3        744.45

sumwx=744.39+248.06+496.36+744.45=2233.26

sumw=3+1+2+3=9

Xbarw= sumwx/sumw

Xbarw= 2233.26/9

Xbarw= 248.14

The weighted mean for line BC is 248.14.

b.

x              w         wx

248.13      1        248.13

248.06     1         248.06

248.18      3        744.54

248.15      3        744.45

sumwx=248.13+248.06+744.54+744.45=1985.18

sumw=3+1+2+3=9

Xbarw= sumwx/sumw

Xbarw= 1985.18/9

Xbarw= 220.576

The weighted mean for line BC is 220.576.

2.2°  

step-by-step explanation:

let θ = the angle of elevation.

tanθ =77/2020

tanθ = 0.038 12

      θ = tan⁻¹0.038 12

      θ = 2.2°

40 x 50 = 2000

step-by-step explanation: