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Basile [38]
3 years ago
12

Ally has taken 5 quizzes this year. Her mean quiz score is 8 out of 10. She takes another quiz and gets a perfect score of 10. W

hat happens to her mean score?
Mathematics
1 answer:
Vlad [161]3 years ago
7 0

Answer:

It goes up

Step-by-step explanation:

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PLZ HELP ME PLZ PLZ I’LL MARK AS BRAINLIESTT!!
agasfer [191]

Answer:

$0.35

Step-by-step explanation:

16.25/1.25=13

4.55/13=0.35

8 0
3 years ago
Is 27/51 equal to 18/34? Explain your reasonings
ludmilkaskok [199]

Answer:

No 27/51 is greater than 18/34.

Step-by-step explanation:

If you simplify 27/51. It'll be 3/7 and if you simplify 18/34 it will be 9/17. Multiply the two fractions but not across. multiply the denominator with the numerator and the other numerator to the denominator to get 63 on 9/17 and 51 on 3/7. So 27/51 > 18/34.

4 0
3 years ago
Answer this question please and thank you
Georgia [21]

0.377 + 5.51 = 5.887

Sorry I can only solve the first one. Could you give me a hint on how to do the second one ?

5 0
3 years ago
Read 2 more answers
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
Which decimial number means the same as 12/100
bixtya [17]

Answer:

0.12

hope this is the answer you are looking for

Step-by-step explanation:

5 0
4 years ago
Read 2 more answers
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