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Vika [28.1K]
2 years ago
10

The camp cook made 3 3/5 pots of chicken soup. Each serving of soup is 2/5 of a pot. How many servings of chicken soup did the c

ook make?
Mathematics
1 answer:
vivado [14]2 years ago
8 0
9 servings is the corect one
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Suppose I have three variables in my dataset called testscr, str, and income. I first execute the command "reg testscr str, r".
IrinaVladis [17]

Answer:

4. It should be less than coefficient on str in the first regression

Step-by-step explanation:

Since the str and income are positively correlated and the coefficient on income in the second regression is positive, the coefficient on str in the second regression therefore should be less than coefficient on str in the first regression.

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3 years ago
Help me pleasee I need to turn it in tonightt
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Student 1 and student 3 are equivalent
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Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

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3 years ago
Does anyone have a discord server for homework help?​
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I do....................................

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Some values of the linear function 'f' are shown in the table what is the value of 3?
Monica [59]
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