the mean of the following frequency dristibution is 50 find the value of p - math - statistics
the mean of the following frequency dristibution is 50 find the
value of p






class
0 - 20
20 - 40
40 - 60
60 - 80
80 - 100


frequency
17
28
32
p
19
class intervals
frequency
(fi)
xi
di = xi - a
fi di


0 - 20
17
10
-40
-680


20 - 40
28
30
-20
-560


40 - 60
32
50 = a
0
0


60 - 80
p
70
20
20p


80 - 100
19
90
40
760


total
Î£
fi = 96 + p


Î£ fi di = 20p -
480


![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/5130830/2013_07_08_13_01_36/mathmlequation8813644107607516415%20png)
the relationship of wavelength and frequency will appear as:using the wavelength range given, 280-320 nm, frequency range will be calculated using the equation above and by converting the values to the proper units [readmore]each wavelength must be in meters:we need to calculate the frequency for both 280 and 320 nm:range of frequency will appear as 1.07x1015 to 9.38x1014 s-1 for p = 1:10
 for q = 1:10
 if p ~= q
 ...
 end
 end
end
without a for loop:
x = rand(10, 10);
index = (1:10) ~= (1:10).';
mean(x(index)) % mean of elements, which are not on main diagonal v = (4/3)πr³ 

dv/dt = 4πr².dr/dt

so, when the diameter is 20 cm we have:

-4π = 4πr².dr/dt

hence, dr/dt = -1/r²

i.e. dr/dt = -1/10² => -1/100

now, d = 2r

so, d(d)/dt = 2.dr/dt

=> d(d)/dt = -1/50 cm/hour

or, -0.2 mm/hour

:)> solution: in a given data, the highest frequency is 41, which lies in the interval 1000-15000.   
 here, l=10000, `f\_(m)=41=26, f\_(2)=16 and h=5000`   
 `"mode""=l+(f\_(m)-f\_(1))/(2f\_(m)-f\_(1)-f\_(2))xxh`   
 `=1000+((41-26)/(2xx41-26-16))xx5000`   
 `=1000+((15)/(82-42))xx5000`   
 `=1000+((15)/(40))xx5000`   
 `=1000+15xx125=10000+1875=11875`   
 hence, the modal income is 11875. hii..

according to me,the cut off will be around 90-95 marks for general category,50-55 marks for obc category,35-40 marks for sc category and 25-30 marks for st category.so it is difficult to be predict now.

please see the table below for previous years trends:-

 

jee main cut off trends (2013-2017)

| s.no. | category | 2017 | 2016 | 2015 | 2014 | 2013 |
| 1 | general | 81 | 100 | 105 | 115 | 113 |
| 2 | obc-ncl | 49 | 70 | 70 | 74 | 70 |
| 3 | sc | 32 | 52 | 50 | 53 | 50 |
| 4 | st | 27 | 48 | 44 | 47 | 45 | a frequency converter, also known as a frequency changer, is a device that takes incoming power, normally 50 or 60 hz and converts it to 400 hz output power. hello




 pgdm (bm)




  





 female




  





 valr - 75, dm - 75, qa - 75




  





 overall - 95




  





 male




  





 valr - 79.46, dm - 80, qa - 90




  





 overall - 96




  





 pgdm (hrm)




  





 female




  





 valr - 80, dm - 70, qa - 67




  





 overall - 93




  





 male




  





 valr - 85, dm - 75, qa - 72




  





 overall - 94




  





 b-schools & cutoff




  





 xlri jamshedpur - 96+ percentile (i) co-ordinates of point p are `((1xx17+2xx(-1))/(1+2),(1xx10+2xx1)/(1+2))` `=((17-8)/3,(10+2)/3)` `=(9/3,12/3)` `=(3,4)` (ii) `op=sqrt((0-3)^2+(0-4)^2)` `op=sqrt(9+16)` `op=sqrt25` `op=5` units (iii) let ab be divided by the point p(0, y) lying on y-axis in the ratio `therefore (0,y)=((kxx17+1xx(-4))/(k+1),(kxx10+1xx1)/(k+1))` `=>(0,y)=((17k-4)/(k+1),(10k+1)/(k+1))` `=>0=(17k-4)/(k+1)` `=>17k-4=0` `=>k=4/17` thus, the ratio in which the y-axis divide the line ab is 4: 17. the value of head loss due to friction using chezy formula with constant c= 60 for a fluid, flowing through a 30 cm diameter and 70 m long pipe. velocity of fluid inside the pipe is 5 m/s. if the value (or total amount) of these two kinds of coins is thirty dollars..
let n = # of nickles, so 500-n = # of dimes.
 
0.05n + 0.10(500-n) = 30.00 expressed the total value; solve for n and you'll then be able to give each count requested.