can anybody help me.
I need c++ functions to calculate the mean and standard deviation of some numbers?
void mean_stddeviation(float d[ ],float%26amp; mean, float%26amp; std_dev)
{
int l = length(d);
mean =0;
for (int i =0; i%26lt;l;i++)
mean +=d(i);
mean /= l;
for (i=0;i%26lt;l;i++)
std_dev += (d(i) - mean )*(d(i)-mean);
std_dev /= l;
std_dev = sqrt(std_dev);
}
Tuesday, July 14, 2009
How do i remove the standard cd player in the corsa c i have got the security keys but still cannot remove?
Check that you have the right keys - they are u shaped and it does take a bit of tugging to get it out
How do i remove the standard cd player in the corsa c i have got the security keys but still cannot remove?
do the same as neil said above, but on some models it's possible to push the unit from behind to help it on it's way out if u can get ur hand up
Reply:Push the keys into the holes each side of the unit as far as they will go, then push the ends of the keys apart (so the points push inward against the sides of the unit), and gently but firmly pull the unit out.
It may require some jiggling to get it free.
Reply:Just requires a bit more muscle....quite a bugger to remove.
Reply:rip it out lol
you will need to go to kw it fin it be £50
Reply:you dont do it yourself!!!!send it in because you will damage the hidden pins that the standard cd player accomodates close to the back of the cage...
Reply:Firstly have you removed the little grub screws (the pins will not fit the holes if not)
Second is it the single din unit (old shape Corsa) or is it the new double din size.
If double din you need to make sure that the pins (keys as you call them) are not bent. They need to go into the holes dead parallel or they will not release the clips. I have seen these really out of shape brand new out of the packet from halfords so you may well need to reshape them quite a bit to get them to fit properly.
If this does not work don't try to force it out, find a friendly car radio or carphone fitter and ask him to remove it
How do i remove the standard cd player in the corsa c i have got the security keys but still cannot remove?
do the same as neil said above, but on some models it's possible to push the unit from behind to help it on it's way out if u can get ur hand up
Reply:Push the keys into the holes each side of the unit as far as they will go, then push the ends of the keys apart (so the points push inward against the sides of the unit), and gently but firmly pull the unit out.
It may require some jiggling to get it free.
Reply:Just requires a bit more muscle....quite a bugger to remove.
Reply:rip it out lol
you will need to go to kw it fin it be £50
Reply:you dont do it yourself!!!!send it in because you will damage the hidden pins that the standard cd player accomodates close to the back of the cage...
Reply:Firstly have you removed the little grub screws (the pins will not fit the holes if not)
Second is it the single din unit (old shape Corsa) or is it the new double din size.
If double din you need to make sure that the pins (keys as you call them) are not bent. They need to go into the holes dead parallel or they will not release the clips. I have seen these really out of shape brand new out of the packet from halfords so you may well need to reshape them quite a bit to get them to fit properly.
If this does not work don't try to force it out, find a friendly car radio or carphone fitter and ask him to remove it
I want to view tenth standard icse result please help name C.M. PONNANNA ROLL NO. - T/1085/047?
www.cisce.indiaresult.com
A gas has a pressure of 0.292 atm at 46.2 ° C. What is its pressure at standard temperature?
0.164 atm
P1/T1 = P2/T2.
Standard temp = 273,15 = T2
T1 = 273,15+ 46.2
P1 = 0.292;
A gas has a pressure of 0.292 atm at 46.2 ° C. What is its pressure at standard temperature?
I have no idea...
P1/T1 = P2/T2.
Standard temp = 273,15 = T2
T1 = 273,15+ 46.2
P1 = 0.292;
A gas has a pressure of 0.292 atm at 46.2 ° C. What is its pressure at standard temperature?
I have no idea...
A gas has a pressure of 0.437 atm at 65.8 °C. What is its pressure at standard temperature?
I presume at constant volume
T1 = 65.8 + 273 = 388.8 K
T2 =%26gt; standard temperature = 273 K
0.437 / 388.8 = p2 / 273
p2 = 0.307 atm
T1 = 65.8 + 273 = 388.8 K
T2 =%26gt; standard temperature = 273 K
0.437 / 388.8 = p2 / 273
p2 = 0.307 atm
A gas has a pressure of 699.0 mmHg at 40.0°C. What is the temperature at standard pressure?
Please use significant figures and show work.
A gas has a pressure of 699.0 mmHg at 40.0°C. What is the temperature at standard pressure?
Gay Lussac's Law. As volume isn't mentioned, I'll assume it's constant.
P1 x T2 = P2 x T1
699mmHg x T2 = 760mmHg x 313K (40°C)
T2 = 760 x 313 ÷ 699
Final Pressure = 237,880 ÷ 699 = 340.3K = 67.3°C.
Reply:Sorry about that but a slight error in my answer. I put 'Final Pressure' instead of Temperature. Report It
Reply:P1= 699.0 mm Hg= 93 kPa
T1= 40.0+273=313 K
P2=101.3 (standart pressure)
T2= X
93 divided by 313= .30X
101.3 divided by .30
temperature= 337 kelvin= 64 degrees celcius
i think this is right...
Reply:699.0 mm Hg / 313 K = 760 mm Hg / x
A gas has a pressure of 699.0 mmHg at 40.0°C. What is the temperature at standard pressure?
Gay Lussac's Law. As volume isn't mentioned, I'll assume it's constant.
P1 x T2 = P2 x T1
699mmHg x T2 = 760mmHg x 313K (40°C)
T2 = 760 x 313 ÷ 699
Final Pressure = 237,880 ÷ 699 = 340.3K = 67.3°C.
Reply:Sorry about that but a slight error in my answer. I put 'Final Pressure' instead of Temperature. Report It
Reply:P1= 699.0 mm Hg= 93 kPa
T1= 40.0+273=313 K
P2=101.3 (standart pressure)
T2= X
93 divided by 313= .30X
101.3 divided by .30
temperature= 337 kelvin= 64 degrees celcius
i think this is right...
Reply:699.0 mm Hg / 313 K = 760 mm Hg / x
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