<span><u>The answer is A. 72.25 percent.</u>
The Hardy-Weinberg principle can be used:</span>
<em>p² + 2pq + q² = 1</em> and <em>p + q = 1</em>
where <em>p</em> and <em>q</em> are the frequencies of
the alleles, and <em>p²</em>, <em>q²</em> and <em>2pq </em>are the
frequencies of the genotypes.
<span>The <em>r</em> allele (<em>q</em>) is found in 15% of the population:
q = 15% = 15/100
Thus, q = </span><span>0.15
To calculate the <em>R</em> allele frequency (<em>p</em>), the formula p + q = 1 is
used:
If p + q = 1, then p = 1 - q
p = 1 - 0.15
Thus, </span><span>p = 0.85
Knowing the frequency of the <em>R</em> allele (<em>p</em>), it is easy to determine the
frequency of the RR genotype (p²):
p² = 0.85² = 0.7225
Expressed in percentage, p² = 72.25%.</span>
Both represent physical weathering.
Answer:
Yes
Explanation:
Range rule of thumb predicts the Range to be a multiple of 4 of the standard deviation or to be four times the standard deviation. Making the usual values equal to 2 standard deviations distanct of the mean of the data distribution.
In a given distribution with mean and standard deviation that is obtained, the usual values in mean (as seen in the attached image).
2*standard deviation and mean + 2*standard deviation.
If the data point is not up to the mean
- 2* standard deviation is taken to be significantly low.
If the data point is more than the mean
+ 2*standard deviation is taken to be significantly high.
Let's take the xbar to be the mean and s as standard deviaiton
Given,
mean, xbar = 1116.2
standard deviation, s =127.7
The range rule of thumb shows that the usual values are within 2 standard deviations from the mean
Lower boundary
= xbar - 2s
= 1116.2 - 2(127.7)
= 860.8
Upper boundary
= xbar + 2s
= 1116.2 + 2(127.7)
= 1371.6
We should note that 1411.6 is not between 860.8 and 1371.6, which connotes that 1411.6cm^3 is unusually high.
I believe it’s A……………………..