Answer:
I believe C is the correct answer I think not a 100% sure
Explanation:
The given question is incomplete. The complete question is as follows;
The number of bacteria in a certain population is predicted to increase according to a continuous exponential growth model, at a relative rate of 16% per hour. Suppose that a sample culture has an initial population of 71 bacteria. Find the predicted population after three hours Do not round any intermediate computations, and round your answer to the nearest tenth bacteria
.
Answer:
114.7
Explanation:
A (t) represent the population of the bacteria at the time t.
Since, the population grows exponentially, the population can be calculated as follows:
A (t) = Ao × 
A (t) is teh final population, Ao is the initial population, e is the exponential, k is rate and t is time.
A (t) = 71 × 
For t = 3 hours
A (t) = 71 × 
A (t) = 114.7.
The population of bacteria after 3 hours is 114.7.
Answer:
Seminal vesicle
Explanation:
Seminal vesicle secretes an alkaline and viscous fluid that constitutes about 60% of the volume of semen. The fluid secreted by seminal vesicle contains fructose which is a monosaccharide sugar, prostaglandins, and clotting proteins. The alkaline nature of the seminal fluid neutralizes the acidic environment of the male urethra and the female reproductive tract. This is required as the acidic conditions inactivate and kill sperm.
The fructose sugar serves a fuel for ATP production by sperm. Prostaglandins are responsible for sperm motility and viability as they stimulate smooth muscle contractions within the female reproductive tract. Therefore, a malfunctioning seminal vesicle would result in low semen volume and reduced motility due to the scarcity of energy.
Answer:
G and K
Explanation:
Crossing-over is a genetic phenomenon that occurs in meiosis, specifically, Prophase I. It is when chromosomal segment (genes) are exchanged between non-sister chromatids of homologous chromosomes. Crossing-over occurs only to genes that are UNLINKED i.e. genes located on different chromosomes or far apart on the same chromosome.
This unlinked genes are said to have the highest RECOMBINATION FREQUENCY. Crossing-over allows genes on the same chromosome but far apart from each other, assort independently, which allows alleles to be recombined on the same chromosome. This phenomenon does not occur to genes that are close on the same chromosome as they will be inherited together as a unit.
In this case, G and K are the farthest apart, hence, they will have the highest RECOMBINATION FREQUENCY i.e. the likelihood for homologous crossing-over to take place during meiosis.
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.