The chef can use the first brand which is 8%, then she can use 3% of the second brand
Answer:
7/4 or 1 3/4
Step-by-step explanation:
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
Answer:
Option D) 0.92
Step-by-step explanation:
We are given the following probability distribution in the question:
x: 0 1 2 3 4 5
P(X): 0.78 0.14 0.03 0.01 0 0.04
We have to find the probability that a randomly selected student will be absent no more than one day.
Thus, we have to evaluate:
![P(x\leq1)\\=P(X=0) + P(X = 1)\\=0.78 + 0.14\\=0.92](https://tex.z-dn.net/?f=P%28x%5Cleq1%29%5C%5C%3DP%28X%3D0%29%20%2B%20P%28X%20%3D%201%29%5C%5C%3D0.78%20%2B%200.14%5C%5C%3D0.92)
0.92 is the probability that a randomly selected student will be absent no more than one day.
Thus, the correct answer is
Option D) 0.92
Answer:
2/3/+3=3
Step-by-step explanation: