The link leads to a single table, and that table doesn't show [ y = 18x ].
You're looking for a table in which each 'y'-value is
18 times the 'x'-value that's right next to it.
Good luck in your quest.
Answer: Work done by the particle is 78 N.
Step-by-step explanation:
If C is the path the particle follows, then work done is 
.
According to the question the force 
and all four points are in the plane 
.
therefore, if S is the at surface with boundary C, so that S is the portion of the plane over the rectangle D=[0,2]\times [0,4].
Now,
By the Stoke's theorem:
![\int_{C}^{}F.dx=\iint_{s}^{}curlF.dS\\\\=\iint_{D}^{}[-8y(0)-2z\left ( \frac{1}{4} \right )+5y]dA\\\\=\iint_{D}^{}\left ( -\frac{1}{2}z+5y \right )dA](https://tex.z-dn.net/?f=%5Cint_%7BC%7D%5E%7B%7DF.dx%3D%5Ciint_%7Bs%7D%5E%7B%7DcurlF.dS%5C%5C%5C%5C%3D%5Ciint_%7BD%7D%5E%7B%7D%5B-8y%280%29-2z%5Cleft%20%28%20%5Cfrac%7B1%7D%7B4%7D%20%5Cright%20%29%2B5y%5DdA%5C%5C%5C%5C%3D%5Ciint_%7BD%7D%5E%7B%7D%5Cleft%20%28%20-%5Cfrac%7B1%7D%7B2%7Dz%2B5y%20%5Cright%20%29dA)
![\int_{0}^{2}\left [ \frac{39}{16}y^2 \right ]_{0}^{4}dx\\\\=\int_{0}^{2}39\, \, dx\\\\=39\times 2\\\\=78 \, N](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B2%7D%5Cleft%20%5B%20%5Cfrac%7B39%7D%7B16%7Dy%5E2%20%5Cright%20%5D_%7B0%7D%5E%7B4%7Ddx%5C%5C%5C%5C%3D%5Cint_%7B0%7D%5E%7B2%7D39%5C%2C%20%5C%2C%20%20dx%5C%5C%5C%5C%3D39%5Ctimes%202%5C%5C%5C%5C%3D78%20%5C%2C%20N)
10 square inch
2 times 5
2+2+2+2+2
The probability of selecting exactly one ace is its likelihood
The probability that a five-card poker hand contains exactly one ace is 29.95%
<h3>How to determine the probability?</h3>
There are 4 aces in a standard deck of 52 cards.
The probability of selecting an ace would be:
p = 4/52
Also, there are 48 non-ace cards in the standard deck
So, the probability of selecting a non-ace after an ace has been selected is:
p = 48/51
The probability of selecting a non-ace up to the fifth selection are:
- After two cards have been selected is: 47/50.
- After three cards have been selected is: 46/49.
- After four cards have been selected is: 45/48.
The required probability is then calculated as:
P(1 Ace) = n * (4/52) * (48/51) * (47/50) * (46/49) * (45/48)
Where n is the number of cards i.e. 5
So, we have:
P(1 Ace) = 5 * (4/52) * (48/51) * (47/50) * (46/49) * (45/48)
Evaluate
P(1 Ace) = 0.2995
Express as percentage
P(1 Ace) = 29.95%
Hence, the probability that a five-card poker hand contains exactly one ace is 29.95%
Read more about probability at:
brainly.com/question/25870256
