Personally, i think this model is wrong because the arrows point that carbon dioxide enters the lungs and leaves as oxygen which is totally the opposite, but he or she has showed that the oxygen breathed in is used in the oxygenation of the blood. I may be wrong but it doesn't seem right if you think about it. But if you look really hard at it, the oxygenated blood changes it to carbon, and then carries it back to the lungs to be inhaled.
Plato One strength of these model is that it clearly shows which gases are taken in and which gases are released for exhalation.
The wingspan is the dependent variable while the bird height is the independent variable.
a) Since R2 is the dependent variable, it can be expressed by the independent variable. An R2 of 93% represents 93% of the variation in the wingspan which can be explained by using the birds height. This means that the appropriateness of the model should not be determined by R2 but by using a scatter plot.
b) The model predicts the value of wingspan using the value of bird height. We cannot say that the bird 10 inches tall has a wingspan if 17 inches instead, we say that the wingspan of the bird which is 10 inches tall is 17 inches.
It is therefore possible that the wing span is not exactly 17 inches
One strength of the model is that it clearly shows which gases are taken in and which gases are released for exhalation.
Your study variable is:
X: Tensile strength of artificial silk fibers. (MJ/m³)
μ= 150 MJ/m³
σ= 10 MJ/m³
Since the only tabulated normal distribution is the standard normal, to obtain the probability that the tensile strength will be between 144.5 and 155.5 MJ/m³ you need to transform the study variable in terms of the standard normal distribution.
To do this transformation, known as standardization, you have to subtract the variable its mean and divide it by its standard deviation.
The first step is to present the probability symbolically:
P(144.5≤X≤155.5)= P(X≤155.5) - P(X≤144.5)
Now you standardize each term using the standard normal Z=(X-μ)/δ
P(Z≤(155.5-150)/10) - P(X≤(144.5-150)/10)
P(Z≤0.55) - P(Z≤-0.55)
The next step is to look for the values in the table to reach the cumulative probability for each of them.
The standard normal table has two sides, the left one shows the accumulated probabilities for negatives values of Z and the right one shows values of accumulated probabilities for positive values of Z. The first column shows the values of the integer plus first decimal of the Z value, the first row shows the second decimal of the Z value, by crossing them both you reach the probability value. (see attachment)
P(Z≤0.55) - P(Z≤-0.55)= 0.7088 - 0.2912= 0.4176
I hope it helps!
the correct answer is option b
behavioral response need communication and thus co-ordination at several levels such as interaction among cells, organ system and the entire organism.
behavioral response produces both internal and environmental stimulus.
it has to be understood that the sea water contains a huge amount of sodium and chlorine ions and if they enter the cell of the atlantic cod fish, the cell will shrivel and the fish will die. to avoid such a condition, the fish uses active transport for the purpose of removing the ions against the gradient.
the atlantic cod is not a maritime osmoconformer. its native application of salts is less than that of the seawater it lives in. it is continually falling water from the gills to the ocean via osmosis.
countering that, the fish is going to be swallowing seawater. but since its internal collection of salts is lower than seawater, the fish will want to go free of excess salt, and since that's going upon a gradient, then active transport, requiring energy, must be used to pump out the salt. much marine fish use the organs to filter out the salt and these fish discharge very straight urine.