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LUCKY_DIMON [66]
3 years ago
8

Create a X and Y table using the expression Y= 4X +3

Mathematics
2 answers:
denpristay [2]3 years ago
6 0
You can use any numbers for the Y you figure out you points. :)

alisha [4.7K]3 years ago
4 0

Answer:

here is the answer

Step-by-step explanation:

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What is the value of x?
butalik [34]

Answer:

it should be 47

Step-by-step explanation:

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WARNING. A USER NAMED MORJHUUTI HAS POSTED A QUESTION SAYING TO JOIN HER VIDEO WITH HER WEBCAM TO WATCH HER MASTER**** PLEASE DO
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ok!!!and no one will be Interested seeing that

4 0
3 years ago
. A referee was standing on a certain yard line as the first quarter ended. He walked 41 3/ 4 yards to a yard line with the same
deff fn [24]
100 yards minus 41 3/4 which is 58 1/4 divided by 2 which is 29 1/2
6 0
3 years ago
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Consider a deck with 2626 black and 2626 red cards. You draw one card at a time and you can choose either guess on whether it is
gregori [183]

The maximum earning is v(r,b)

Let v(r,b) be the expected value of the game for the player, assuming optimal play, if the remaining deck has r red cards and b black cards.

Then v(r,b) satisfies the recursion

and

The stopping rule is simple: Stop when v(r,b)=0.

To explain the recursion . . .

If r,b>0, and the player elects to play a card, then:

  • The revealed card is red with probability \frac{r}{r+b}, and in that case, the player gets a score of +1, and the new value is V(r-1,b)
  • The revealed card is black with probability \frac{b}{r+b}, and in that case, the player gets a score of −1, and the new value is V(r,b-1)

Thus, if r,b>0, electing to play a card yields the value f(r,b).

But the player always has the option to quit, hence, if r,b>0, we get v(r,b)=max(0,f(r,b)).

Implementing the recursion in Maple, the value of the game is

v(26,26)=41984711742427/15997372030584

v(26,26)  ≈2.624475549

and the optimal stopping strategy is as follows . . .

  • If 24≤b≤26, play while r≥b−5.
  • If 17≤b≤23, play while r≥b−4.
  • If 11≤b≤16, play while r≥b−3.
  • If 6≤b≤10, play while r≥b−2.
  • If 3≤b≤5, play while r≥b−1.
  • If 1≤b≤2, play while r≥b.
  • If b=0, play while r>0.  

So, The maximum earning is v(r,b)

Learn more about PROBABILITY here

brainly.com/question/24756209

#SPJ4

4 0
2 years ago
In a previous exercise we formulated a model for learning in the form of the differential equation dp dt = k(m − p) where p(t) m
DaniilM [7]
I assume you mean

   \dfrac{dP}{dt} = k(M-P)

ANSWER
An expression for P(t) is

   
P = M - Me^{-kt}

EXPLANATION
This is a separable differential equation. Treat M and k as constants. Then we can divide both sides by M - P to get the P term with the differential dP and multiply both sides by dt to separate dt from the P terms

   \begin{aligned} \dfrac{dP}{dt} &= k(M-P) \\ \dfrac{dP}{M-P} &= k\, dt
\end{aligned}

Integrate both sides of the equation.

   \begin{aligned}
\int \dfrac{dP}{M-P} &= \int k\, dt \\
-\ln|M-P| &= kt + C \\
\ln|M-P| &= -kt - C\end{aligned}

Note that for the left-hand side, u-substitution gives us 

   u = M - P \implies  du = -1dP \implies dP = -du

hence why \int \frac{dP}{M-P} \ne \ln|M - P|

Now we use the definition of the logarithm to convert into exponential form.

The definition is 

   \ln(a) = b \iff \log_e(a) = b \iff e^b = a

so applying it here, we get

   \begin{aligned} \ln|M-P| &= -kt - C \\ |M - P| &= e^{-kt - C} \\ 
M - P &= \pm e^{-kt - C} 
 \end{aligned}

Exponent properties can be used to address the constant C. We use x^{a} \cdot x^{b} = x^{a+b} here:

   \begin{aligned}
 M - P &= \pm e^{-kt - C} \\
M - P &= \pm e^{- C - kt} \\ 
M - P &= \pm e^{- C + (- kt)} \\ 
M - P &= \pm e^{- C} \cdot e^{- kt} \\ 
M - P &= Ke^{- kt} && (\text{\footnotesize Let $K = \pm e^{-kt}$ }) \\ 
M &= Ke^{- kt} + P\\
P &= M - Ke^{- kt}
\end{aligned}

If we assume that P(0) = 0, then set t = 0 and P = 0

   \begin{aligned} 
0 &= M - Ke^{- k\cdot 0} \\
0 &= M - K \cdot 1 \\
M &= K
 \end{aligned}


Substituting into our original equation, we get our final answer of

   P = M - Me^{-kt}
6 0
4 years ago
Read 2 more answers
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