Answer:
True
Explanation:
Animals can be categorized into 3 based on body symmetry
- <em>Those without any body symmetry (asymmetrical)</em>
- <em>Those with bilateral body symmetry (bilateria)</em>
- <em>Those with radial body symmetry (Radiata)</em>
Animals can be categorized into 2 based on number of embryonic germ layer;
- <em>Those with two layers - endoderm and ectoderm (diplobastic)</em>
- <em>Those with three layers - mesoderm in addition to ectoderm and endoderm (triploblastic)</em>
Animals can be categorized based on presence/absence of body cavity or coelom;
- <em>No body cavity - acoelomates</em>
- <em>False body cavity - pseudocoelomates</em>
- <em>True body cavity - coelomates</em>
Animals can be categorized into 2 based on characteristics of embryonic development;
- <em>Deuterostomes</em>
- <em>Protosomes</em>
Answer:
A) 48
B) 96
C) 48
D) 48
Explanation:
Attached is a table summarizing the number of chromosomes and chromatids in the different stages of mitosis and meiosis in humans who are described as 2n = 46.
For the organism which is described as 2n = 48, substitute 46 in the table for 48 to get the appropriate figures.
8-10 years but one has lived to 20 years.
Answer:
both could be correct
Explanation:
It usually depends on how fast or how severe the change is, if its mild or happenes slowly over time they may adapt, if it is extreame or happenes very quickly the species could very well die out.
Answer:
2.275% of this population has a diastolic blood pressure less than 60 mmHg
Explanation:
Hello!
Yo have the distribution of the diastolic blood pressure in a certain population. Be X: diastolic blood pressure of an individual, X~N(μ;δ²)
Where
μ= 82mmHg
δ=11 mmHg
You need to calculate the probability of an individual of this population having less than 60mmHg diastolic blood pressure.
Symbolically:
P(X<60)
To obtain the value of probability you need to standardize the value of diastolic pressure so that you can obtain it from the standard normal distribution. The way to standardize the value is to subtract the mean and divide by the standard deviation
Z= (X-μ)/δ~N(0;1)
P(Z<(60-82)/11)
P(Z<-2)= 0.02275
I hope it helps!