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If 4% of males in a population have red-green color blindness, then the allelic frequency is 4% in males and in females. If mating is random, then in females, 92.16% do not carry the allele on either X chromosome, 7.68% carry the allele on one X chromosome, and 0.16% carry the allele on both X chromosomes. We have the dominant allele with frequency p and the recessive allele with frequency q, so 0.9216 + 0.0768 + 0.0016 = p^2 + 2pq + q^2 = (p + q)^2 = (0.96 + 0.04)^2
there are no low frequency x-rays
For any wave:wavelength x frequency = speedSpeed of light, in this case.For any wave:wavelength x frequency = speedSpeed of light, in this case.For any wave:wavelength x frequency = speedSpeed of light, in this case.For any wave:wavelength x frequency = speedSpeed of light, in this case.
An X-ray is a high frequency (high energy) electromagnetic wave. It is much higher in frequency than any broadcast wave. The only electromagnetic radiation with higher frequency (shorter wavelength) than the X-ray is the gamma ray.
X-ray has a higher frequency than visible light.
If 4% of males in a population have red-green color blindness, then the allelic frequency is 4% in males and in females. If mating is random, then in females, 92.16% do not carry the allele on either X chromosome, 7.68% carry the allele on one X chromosome, and 0.16% carry the allele on both X chromosomes. We have the dominant allele with frequency p and the recessive allele with frequency q, so 0.9216 + 0.0768 + 0.0016 = p^2 + 2pq + q^2 = (p + q)^2 = (0.96 + 0.04)^2
What allelic frequency will generate twice as many recessive homozygotes as heterozygotes? Answer: We need to solve for the following equation: q2 (aa) = 2 x the frequency of Aa. Thus, q2 (aa) = 2(2pq). Or another way of writing it is q2 = 4 x p x q. We only want q, so lets trash p. Since p = 1 - q, we can substitute 1 - q for p and, thus, q2 = 4 (1 - q) q. Then, if we multiply everything on the right by that lone q, we get q2 = 4q - 4q2. We then divide both sides through by q and get q = 4 - 4q. Subtracting 4 from both sides, and then q (i.e. -4q minus q = -5q) also from both sides, we get -4 = -5q. We then divide through by -5 to get -4/-5 = q, or anotherwards the answer which is 0.8 =q. I cannot imagine you getting this type of problem in this general biology course although if you take algebra good luck.
Main function of Variable Frequency Drive is to be able to run the AC Induction motors at different speeds (by changing the frequency). The speed is a function of frequency. N - Speed. F - Frequency P - Pole N= ( 120 x F)/ P
change that equation to: --P-- = -17 100------92 then cross multiply 17 X 100 = 1700 92 X P= (92)(P) then divide both 1700 and (92)(X) by 92 92 X P----------------1700 ______ _____ = P---------____ = l18.48%l --92---------------------92 _______
6. all you have to do is use the IS over OF strategy. IS 20% and OF 30 so: P 20 30 100 then cross multiplication: P x 100 = 30 x 20 P x 100/100 = 600/100 P = 6 So there you have it, P = 6
speed (of the wave) = frequency x wavelengthspeed (of the wave) = frequency x wavelengthspeed (of the wave) = frequency x wavelengthspeed (of the wave) = frequency x wavelength
6. all you have to do is use the IS over OF strategy. IS 20% and OF 30 so: P 20 30 100 then cross multiplication: P x 100 = 30 x 20 P x 100/100 = 600/100 P = 6 So there you have it, P = 6
change that equation to: --P-- = -17 100------92 then cross multiply 17 X 100 = 1700 92 X P= (92)(P) then divide both 1700 and (92)(X) by 92 92 X P----------------1700 ______ _____ = P---------____ = l18.48%l --92---------------------92 _______
The formula you are looking for is f = N x P/120. (f) generator Frequency = (N) number of revolutions per minute of the engine x (P) number of magnetic poles / 120 2400 x 2 / 120 = 40 Hz
Use the relation: speed = frequency x wavelengthUse the relation: speed = frequency x wavelengthUse the relation: speed = frequency x wavelengthUse the relation: speed = frequency x wavelength
The cumulative frequency of a random variable X for the value x is the number of observations such that X ? x.
If you start with a fraction p/q and are told that x/y is an equivalent fraction, then the simplest check is to cross-multiply: p*y must be equal to q*x.