Q: What does 32 p on a cb mean?

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126.5that should be the socket where you plug in the mike.

p = 8

Depending on what number p represents, it is 8xp. If p = 4, then 8x4=32.

A=s2 64=s2 s=8 units P=4s P=4(8) P=32 units Therefore, the perimetre is 32 units.

The equation is as follows: p - 5/8 (1/4 + p). This equation expands to (1/4)p + p^2 - 5/32 - (5/8)p, which simplifies to p^2 - (3/8)p - 5/32. Now either solve for p using the quadratic formula or solve by factoring.

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126.5that should be the socket where you plug in the mike.

I'm currently searching 35 p on the mw.... one thought led me to the millennium wheel, on which there are 32, not 35, pods, which may be your answer??

p + 32

4p = 32 Therefore, p = 32/4 p = 8

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No. 32 ACP vs. 380 ACP

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Assuming each toss is independent, you can use the binomial distribution,P( X = 32) = 50C32*(p)32*(1-p)50-32 where p is the probablity of getting heads on a single toss. Assuming that the coin is fair, p = 1/2.So the answer is 50C32*(1/2)50 = 50!/(32!*18!*250 = 0.016 approx.Assuming each toss is independent, you can use the binomial distribution,P( X = 32) = 50C32*(p)32*(1-p)50-32 where p is the probablity of getting heads on a single toss. Assuming that the coin is fair, p = 1/2.So the answer is 50C32*(1/2)50 = 50!/(32!*18!*250 = 0.016 approx.Assuming each toss is independent, you can use the binomial distribution,P( X = 32) = 50C32*(p)32*(1-p)50-32 where p is the probablity of getting heads on a single toss. Assuming that the coin is fair, p = 1/2.So the answer is 50C32*(1/2)50 = 50!/(32!*18!*250 = 0.016 approx.Assuming each toss is independent, you can use the binomial distribution,P( X = 32) = 50C32*(p)32*(1-p)50-32 where p is the probablity of getting heads on a single toss. Assuming that the coin is fair, p = 1/2.So the answer is 50C32*(1/2)50 = 50!/(32!*18!*250 = 0.016 approx.

32 pieces on a chess

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