To find the probability ( p ) of randomly selecting a green organism from the group, you divide the number of green organisms by the total number of organisms. In this case, ( p = \frac{28}{100} = 0.28 ). Therefore, the probability ( p ) of selecting a green organism is 0.28, or 28%.
The answer depends on the definition of p.
If 12 out of 100 organisms have short legs, P = 0.65.
Without replacement: P(one green and one blue) = P(drawing green then blue) + P(drawing blue then green) = (6/23)(9/22) + (9/23)(6/22) = 104/506 = 52/253 With replacement: P(one green and one blue) = P(green)*P(blue) = (6/23)(9/23) = 54/529
Hold on to your hat! Suppose the rate of depreciation is p percent per year. Current value = Start Value*(1 - percentage/100)years That gives 2047.08 = 5500*(1 - p/100)9 So 0.3722 = (1 - p/100)9 or log(0.3722) = 9*log(1-p/100) -0.4292 = 9*log(1-p/100) -0.0477 = log(1-p/100) 10-0.0477 = 1-p/100 0.8960 = 1-p/100 p/100 = 1 - 0.8960 = 0.1040 and, finally, p = 0.1040*100 = 10.4%
P over B equals R over 100 => P/B = R/100 => P/5950 = 48000/100 => P = 5950 * 480 = 2856000 Or 59.50 * 48000 = 2856000
If the green allele is recessive and there are 28 organisms with green eyes, then the frequency of the green allele (q) would be √(28/Total organisms) = √(28/Total organisms). Since p+q=1, p = 1 - q. Substituting 28 for q gives 1 - √(28/Total organisms).
The answer depends on the definition of p.
, 0.34 Apex
If 59 out of 100 organisms are P, then the remaining organisms would be Q. Therefore, Q would be 41 out of 100 organisms.
If 12 out of 100 organisms have short legs, P = 0.65.
.54
To find q (the frequency of the green allele), use the Hardy-Weinberg equation: p^2 + 2pq + q^2 = 1. Given that 11% are green, q^2 = 0.11. Therefore, q = √0.11 ≈ 0.33.
0.16
There are 100 p in a pound so 28p as a fraction of a pound = 28/100 = 7/25
If 12 out of 100 organisms have short legs, this represents the homozygous recessive genotype (qq). Therefore, the frequency of the recessive genotype (q²) is 0.12. To find q, we take the square root of 0.12, which gives q ≈ 0.346. Since p + q = 1, we can calculate p as 1 - q, resulting in p ≈ 0.654.
A. 0.45 Apex
To find the frequency of the recessive allele (q) in the fly population, we can use the Hardy-Weinberg principle. If 120 out of 100 organisms are green, this indicates that 120 are homozygous recessive (qq). Assuming the total population is 120, the frequency of the homozygous recessive genotype (qq) is 120/120 = 1. Since q² = 1, we find that q = 1. Thus, the frequency of the recessive allele (q) is 1, meaning all individuals have green eyes.