Question

plzzzzz help me out... ​

Answer 1

∆ Diagram:- Refer the attachment

Given:-

Height of first pole = 8m

Height of second pole = 16m

Distance between the feet of poles = 15m

To Find:-

The distance between the tops of the two poles.

Solution:-

Let the height of the first pole be AB = 8m

And height of the season pole be DC = 16m

• AC = 15m

Let us draw a line BE perpendicular to CD. i.e BE ⊥CD

And, AC ⊥ DC

So, BE = AC = 15m

And, AB = EC = 8m

Now:-

$\sf \implies DE = DC - EC$

$\sf \implies DE = 16 - 8$

$\sf \implies DE = 8m$

$\sf In \: \triangle BED$

By applying Pythagoras Theorem:-

$\sf \implies (Hypotenuse)^2 = (Base)^2 + (Height)^2$

$\sf \implies BD)^2 = DE^2 + (BE)^2$

$\sf \implies BD^2 = (8)^2 + (15)^2$

$\sf \implies BD^2 = 64 + 225$

$\sf \implies BD^2 = 289$

$\sf \implies BD = \implies{289}$

$\sf \implies BD = 17m$

$\underline{\boxed{\therefore \textsf{\textbf{Distance \: between \: their \: tops = 17m}}}}$

Answer 2

Answer:

The mean of the daily wages of 50 workers is 545.2 Rs.

Step-by-step-explanation:

We have given the frequency distribution of the daily wages of 50 workers.

We have to find the mean of the daily wages of those workers.

Now,

$\begin{gathered}\displaystyle{\begin{array}{|c|c|c|c|}\cline{1-4}\bf\:Class\:(\:Daily\:wages\:in\:Rs\:) & \bf\:Class\:mark\:(\:x_i\:) & \bf\:Frequency\:(\:f_i\:) & \bf\:x_i\:f_i\\\cline{1-4}\sf\:500\:-\:520 & \sf\:510 & \sf\:12 & \sf\:6120\\\cline{1-4}\sf\:520\:-\:540 & \sf\:530 & \sf\:14 & \sf\:7420\\\cline{1-4}\sf\:540\:-\:560 & \sf\:550 & \sf\:8 & \sf\:4400\\\cline{1-4}\sf\:560\:-\:580 & \sf\:570 & \sf\:6 & \sf\:3420\\\cline{1-4}\sf\:580\:-\:600 & \sf\:590 & \sf\:10 & \sf\:5900\\\cline{1-4}& & \sf\:N\:=\:\sum\:f_i\:=\:50 & \sf\:\sum\:x_i\:f_i\:=\:27260\\\cline{1-4}\end{array}}\end{gathered}$

Now,

$\displaystyle{\pink{\sf\:Mean\:\overline{X}\:=\:\dfrac{\sum\:x_i\:f_i}{\sum\:f_i}}\sf\:\:\:-\:-\:-\:[\:Formula\:]}$

$\displaystyle{\implies\sf\:Mean\:\overline{X}\:=\:\cancel{\dfrac{27260}{50}}}$

$\displaystyle{\implies\sf\:Mean\:\overline{X}\:=\:\cancel{\dfrac{13630}{25}}}$

$\displaystyle{\implies\sf\:Mean\:\overline{X}\:=\:\cancel{\dfrac{2726}{5}}}$

$\displaystyle{\implies\boxed{\red{\sf\:Mean\:\overline{X}\:=\:545.2\:Rs}}}$

The mean of the daily wages of 50 workers is 545.2 Rs.