Jacob Northover retweetledi
Jacob Northover
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Jacob Northover
@JacobNorthover
Southampton, England Katılım Ağustos 2016
288 Takip Edilen86 Takipçiler
Jacob Northover retweetledi
🚨 NEW: #thfc are genuinely interested in Fiorentina midfielder Nicolò Fagioli.
He is one of the first names on Spurs' list for the summer. | @marcoconterio
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RT @David_Ornstein: 🚨 EXCLUSIVE: Thomas Frank sacked as Tottenham Hotspur head coach. #THFC working through various possible contingency pl…
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Tottenham are worried Mohammed Kudus deal will get hijacked, and with good reason
spurs-web.com/spurs-news/tot…
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I can't believe how many people I'd seen call Solanke a flop online in recent months.
His assist against Konate was proper CF play—few could manage that. With 17 G/A this season, plus his off-the-ball work rate, pressing, ability to drop deep, lead the line, and hold off opponents, he's been outstanding. And let's not forget, he's had no rest because of Richarlison's injury.
Also I know it was offside, but the run and finish earlier also so good.
Anyone who called him a flop has clearly never watched him play this season.
#Tottenham

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Got interested in statistics - began trying to extrapolate the amount of minutes Mikey Moore will play tonight.
First 5 Tottenham games, he played 0 minutes. This simply a line of y=0.
Next 3 Tottenham games, he played 2 minutes, followed by 5, followed by 14.
So simply, we can create an exponential growth model by using the coordinates (5,2), (6,5), and (7,14).
We can use the equation y=ae^(b(x−4))+c to figure out an appropriate exponential equation. We have to use x-4 to account for the fact that this growth starts after the 5th game (or here, the y=4 coordinate, since we started graphing at 0).
Using a system of equations, we can substitute each coordinates into the y and x to get 3 different equations:
1. 2=ae^b+c
2. 5=ae^(2b)+c
3. 14=ae^(3b)+c
From here, we do the simple solving for C by equating 2 of the equations to each other. We then subtract the 1st equation from the 2nd to get 3=a(e^b)((e^b)−1), and the 3rd from the 2nd to get 9=a(e^2b)((e^b)−1).
Since we have successfully eliminated c, we need to eliminate one more variable, and we can eliminate a by dividing the 2nd equation we got by the 1st.
This will leave us with 9/3 = a(e^2b)((e^b)−1)/3=a(e^b)((e^b)−1). First half of the equation is smiply 3, the first term of the 2nd half will become e^b as the a term disappears, whilst we subtract the indicies. The 2nd term of the 2nd half will also disappear.
So we 3=e^b. And we can use natural logarithms to solve for b, meaning b is simply ln3.
Now is the easy the part, we have b, so we can just substitute ln3 into the first equation that we got via subtraction. 3=a(e^b)(e^b−1) which becomes 3=a(e^ln3)((e^ln3)-1). e to the ln3 is simply just 3, so this becomes easy. 3=a(3)(2) so we have 3=6a which means a = 1/2.
Finally, we can work out c, which is done by substituting b and a into the original equations. Any of them. To save you boredom, I will tell you I arrived at c=1/2.
Now we can substitue our values to get a value for the curve that we can use to extrapolate Mikey Moore's minutes. y=1/2(e^((ln3))(x-4)))+1/2, which can be simplified to y=1/2(3^(x-4))+1/2.
So, I decided to plot this on Desmos.
Based on extrapolation, I predict Mikey Moore to see 41 minutes vs. Ferencvaros.

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