Answer:
The inference that can be made using the dot plot is:
The range of round 1 is greater than the round 2 range.
Step-by-step explanation:
<u>Round 1:</u>
Score Frequency
1 0
2 2
3 3
4 2
5 1
Hence, the minimum score of Round 1 is: 2
maximum score is: 5
Hence, Range=Maximum value-Minimum score
=5-2
=3
Similarly, <u>Round-2</u>
Score Frequency
1 0
2 0
3 0
4 4
5 4
Hence, the minimum score of Round 1 is: 4
maximum score is: 5
Hence, Range=Maximum value-Minimum score
=5-4
=1
The scores of round 2 are higher than round-1.
Since round 2 have a higher frequency for higher scores as compared to round-1.
Hence, Range of round 1 is greater than the range of Round-2.
Answer:
the last choice
Step-by-step explanation:
two functions are said to be inverse when they are symmetric about the line y=x
It looks like the vector field is
<em>F</em><em>(x, y)</em> = 3<em>x</em> ^(2/3) <em>i</em> + <em>e</em> ^(<em>y</em>/5) <em>j</em>
<em></em>
Find a scalar function <em>f</em> such that grad <em>f</em> = <em>F</em> :
∂<em>f</em>/∂<em>x</em> = 3<em>x</em> ^(2/3) => <em>f(x, y)</em> = 9/5 <em>x</em> ^(5/3) + <em>g(y)</em>
=> ∂<em>f</em>/∂<em>y</em> = <em>e</em> ^(<em>y</em>/5) = d<em>g</em>/d<em>y</em> => <em>g(y)</em> = 5<em>e</em> ^(<em>y</em>/5) + <em>K</em>
=> <em>f(x, y)</em> = 9/5 <em>x</em> ^(5/3) + 5<em>e</em> ^(<em>y</em>/5) + <em>K</em>
(where <em>K</em> is an arbitrary constant)
By the fundamental theorem, the integral of <em>F</em> over the given path is
∫<em>c</em> <em>F</em> • d<em>r</em> = <em>f</em> (0, 1) - <em>f</em> (1, 0) = 5<em>e</em> ^(1/5) - 34/5
