The answer depends on the definition of p.
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∙ 7y agoIf 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 12 out of 100 organisms have short legs, P = 0.65.
, 0.34 Apex
A. 0.45 Apex
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.
If 59 out of 100 organisms are P, then the remaining organisms would be Q. Therefore, Q would be 41 out of 100 organisms.
.54
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).
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.
Timothy P. Green was born in 1963.
Jay P. Green was born in 1918.
Jay P. Green died in 2008.