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Ierofanga [76]
3 years ago
14

Lin plays a game that involves a standard number cube and a deck of ten cards numbered 1 through 10. If both the cube and card h

ave the same number, Lin gets another turn. Otherwise, play continues with the next player. What is the probability that Lin gets another turn?
Mathematics
1 answer:
fomenos3 years ago
4 0
1/20, 1 in 20 turns,...............
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Can someone help me and show your work thanks
qwelly [4]

Answer:

10.08

Step-by-step explanation:

7 0
3 years ago
What is the difference of the whole number and the mixed number with an improper fraction below? 8 7/4 -7 A. 2 1/2 B. 2 3/4 C. 4
MatroZZZ [7]

Answer:

D. 2 1/4

Step-by-step explanation:

8 7/4 - 7

9 3/4 - 7

2 1/4

6 0
3 years ago
Use lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y = xyz; x^
Snezhnost [94]
I'm assuming the constraint involves some plus signs that aren't appearing for some reason, so that you're finding the extrema subject to x^2+2y^2+3z^2=96.

Set f(x,y,z)=xyz and g(x,y,z)=x^2+2y^2+3z^2-96, so that the Lagrangian is

L(x,y,z,\lambda)=xyz+\lambda(x^2+2y^2+3z^2-96)

Take the partial derivatives and set them equal to zero.

\begin{cases}L_x=yz+2\lambda x=0\\L_y=xz+4\lambda y=0\\L_z=xy+6\lambda z=0\\L_\lambda=x^2+2y^2+3z^2-96=0\end{cases}

One way to find the possible critical points is to multiply the first three equations by the variable that is missing in the first term and dividing by 2. This gives

\begin{cases}\dfrac{xyz}2+\lambda x^2=0\\\\\dfrac{xyz}2+2\lambda y^2=0\\\\\dfrac{xyz}2+3\lambda z^2=0\\\\x^2+2y^2+3y^2=96\end{cases}

So by adding the first three equations together, you end up with

\dfrac32xyz+\lambda(x^2+2y^2+3z^2)=0

and the fourth equation allows you to write

\dfrac32xyz+96\lambda=0\implies \dfrac{xyz}2=-32\lambda

Now, substituting this into the first three equations in the most recent system yields

\begin{cases}-32\lambda+\lambda x^2=0\\-32\lambda+2\lambda y^2=0\\-32\lambda+3\lambda z^2=0\end{cases}\implies\begin{cases}x=\pm4\sqrt2\\y=\pm4\\z=\pm4\sqrt{\dfrac23}\end{cases}

So we found a grand total of 8 possible critical points. Evaluating f(x,y,z)=xyz at each of these points, you find that f(x,y,z) attains a maximum value of \dfrac{128}{\sqrt3} whenever exactly none or two of the critical points' coordinates are negative (four cases of this), and a minimum value of -\dfrac{128}{\sqrt3} whenever exactly one or all of the critical points' coordinates are negative.
6 0
3 years ago
Michael had $550 in his savings account. He took out $30.75 every month for one year. What is the net change in Michael’s accoun
gulaghasi [49]

Answer:

B

Step-by-step explanation:

5 0
4 years ago
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