It may be 41.
To find the frequency of the "r" allele (q) in the population, divide the number of red organisms by the total number of organisms: 59/100 = 0.59. So, q = 0.59.
.77
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).
.93
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.
The question seems to be missing context regarding what "q" represents. If "q" refers to a proportion or percentage of red organisms, you could express it as a fraction of the total: for 59 red organisms out of 100, q would be 0.59 or 59%. If all 100 organisms are red, then q would be 1 or 100%. Please clarify if you need a different interpretation.
To find the frequency of the "r" allele (q) in the population, divide the number of red organisms by the total number of organisms: 59/100 = 0.59. So, q = 0.59.
If 30 out of 100 organisms are red, then the proportion of red organisms can be represented as ( q = \frac{30}{100} = 0.3 ) or 30%. Therefore, ( q ) is 0.3.
.77
For this problem, assume q is 100. So, if p is 40 percent, that would mean 40/100 which equals .4 or 40 percent. So, 100/40 equal 2.5 or 250 percent. If p is 40 percent of q, then q is 250 percent of p.
Suppose the value of whatever it is, is P in the first year and Q in the next. Then the percentage change is 100*(P - Q)/Q or, equivalently, 100*(P/Q - 1)
The quantity, Q, demanded at price P is 100 - 4Q So Q = 25 - P/4 And therefore, the demand elasticity is -1/4 or -0.25, whatever the value of Q.
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).
, 0.34 Apex
int *p, *q; p = (int*) malloc (100 * sizeof (int)); q = (int*) calloc (100, sizeof (int)); Note that p is left in an uninitialised state whereas q is initialised with the value zero.
n/100 is nice since it can be expressed as a n percent. The goal is for your fraction p/q, can find out what q must be multiplied by to get 100 and then multiply that by p also. Many times it will not be this simple since for starters, q may be >100 then we divide, or q may not be a divisor of 100 and there are many other pitfalls.
with two figures dive the smaller by the bigger and times by 100