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 you mean: p/35.60 = 18/100 then by multiplying both sides by 35.60 the value of p works out as 6.408
0.16
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.
, 0.34 Apex
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).
.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.
A. 0.45 Apex
Timothy P. Green was born in 1963.
Jay P. Green was born in 1918.
Jay P. Green died in 2008.
72 p/1 pound = 72 p /100 p = 72/100 = 18/25, in its simplest form.72 p/1 pound = 72 p /100 p = 72/100 = 18/25, in its simplest form.72 p/1 pound = 72 p /100 p = 72/100 = 18/25, in its simplest form.72 p/1 pound = 72 p /100 p = 72/100 = 18/25, in its simplest form.