Question set 2: Many species of terrestrial tree frogs that hibernate at or near the ground surface can survive prolonged exposu
re to low winter temperatures. In freezing conditions, the frog’s body temperature, called its supercooling temperature, remains relatively higher because of an accumulation of glycerol in its body fluids. A study in Science revealed that the supercooling temperature (in Celsius scale,) of terrestrial frogs frozen at -6°_C has a relative frequency distribution with a mean of -2°_C and a standard deviation of 0.3°_C. Consider the mean supercooling temperature of a random sample of n = 42 terrestrial frogs frozen at -6°_C. (a) Find the probability that the mean of the sample exceeds -2.05°_C .
1 answer:
Answer:
0.86
Step-by-step explanation:
The z-score we need to make the comparison is given by
where
= the standard deviation established in the study
n = the sample size
and for the Normal distribution N(0,1) we want to find
<em>P(X > -1.0801) </em>
By using a spreadsheet or the table attached, we find that
P(X > -1.0801) = 0.86
(See picture attached)
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Answer:
1)f(x)=2(3)^x
2)f(x)=10(2)^x
3)Graph attached
4) Amber will have 192 and 3145728 shares on day 3rd and 10th respectively
Ben will have 54 and 118,098 shares on day 3rd and 10th respectively
Carter will have 80 and 10240 shares on day 3rd and 10th respectively
5)the graph has moved 45 unit up
6)Amber's posts will travel fastest
7)Amber' function
Step-by-step explanation:
1)
Ben shares posts with two friends and then they share with 3 more everyday, so her the growth rate is 3
f(x)=2(3)^x
x=number of days
2)
Carter shares posts with ten friends and then they share with 2 more everyday, so her the growth rate is 2
f(x)=10(2)^x
x=number of days
3)
Graph attached contain graph's of following
Ben's f(x)=2(3)^x in red
Carter's f(x)=10(2)^x in blue
Amber's f(x)=3(4)^x in green
4)
Finding how many shares each student will get on day 3 and day 10:
For Amber:
substituting x=3,
f(x)=3(4)^3
= 192
Now substituting x=10,
f(x)= 3(4)^10
= 3145728
Amber will have 192 and 3145728 shares on day 3rd and 10th respectively
For Ben:
substituting x=3,
f(x)=2(3)^3
= 54
Now substituting x=10,
f(x)= 2(3)^10
= 118,098
Ben will have 54 and 118,098 shares on day 3rd and 10th respectively
For Carter:
substituting x=3,
f(x)=10(2)^3
= 80
Now substituting x=10,
f(x)= 10(2)^10
= 10240
Carter will have 80 and 10240 shares on day 3rd and 10th respectively
5)
If Amber's graph changes to
f'(x)=3(4)^x +45
then as per translation, it means the graph has moved 45 unit up from original position as shown in attached graph
original f(x)=3(4)^x in green
translated f'(x)=3(4)^x +45 in blue
6)
In exponential function the base of function effects the growth and in given cases of Amber, Ben and Carter; base of Amber's function is greatest i.e 4 other two have 2 and 3.
So Amber's posts will travel fastest
7) A post with fewer friends initially and then more shares like Amber. Because at the end of the day the growth rate is important and as explain above the number of shares i.e base of exponential function is important.
Hence for more shares and likes Amber' function is the best choice !
