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3 years ago

MATH HELP ASAP! PLEASE HELP! 20 POINTS!!

Mathematics

1 answer:

Aloiza [94]3 years ago

8 0

The mean decreases, think about it, if every single value in a data set was decreased by the same number, the average would also go down by the same number

The median would also go down, because the value in the middle of the data set would now be smaller than before

The mode would also decrease, because the value that showed up the most in the data set has now also been decreased

The range, however, stays the same. If the lowest value in a data set was 3 and the highest was 10 and you subtracted 2 from each value in the data set, you would have the lowest at 1 and the highest at 8. But, 10-3 still equals 8-1, so the range is unchanged

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What is overlapping set

Westkost [7]

Answer:

two sets are said to be overlapping if they contain at least one element in common A=(1,2,3,4)and B =(4,7,1,9)are said to be overlapping sets.

Step-by-step explanation:

please answer my question I answer your question

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6 0

3 years ago

A cereal box can hold 144 cubic inches of of cereal. Suppose the box is 8 inches long and

Korvikt [17]

$Answer: \\ Volume \: of \: the \: box:V = lwh \\ \Rightarrow h = \frac{V}{lw} = \frac{144}{8 \times 1.5} = \frac{144}{12} = 12 \\ \Rightarrow The \: box's \: height:12 \: inches$

5 0

3 years ago

If p(x) = x2 – 1 and 4(x) = 5(x-1), which expression is equivalent to (p -9)(x)?

jenyasd209 [6]

Answer:

5 times 5 24

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Jordan has gone into the nut business. He wants to make a mixture of walnuts and cashews that cost $1.00 per pound. How many pou

vampirchik [111]

Answer:

b.) 11 Ibs.

Step-by-step explanation:

6 0

3 years ago

Find the critical points of the function f(x, y) = 8y2x − 8yx2 + 9xy. Determine whether they are local minima, local maxima, or

NARA [144]

Answer:

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

Step-by-step explanation:

The function is:

$f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y$

The partial derivatives of the function are included below:

$\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y$

$\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)$

$\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x$

$\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)$

Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.

$y \cdot (8\cdot y -16\cdot x + 9) = 0$

$x \cdot (16\cdot y - 8\cdot x + 9) = 0$

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

$\left \{ {{-16\cdot x + 8\cdot y=-9} \atop {-8\cdot x + 16\cdot y=-9}} \right.$

The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

$x\cdot (-8\cdot x + 9) = 0$

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

$y\cdot (8\cdot y+9) = 0$

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

$H = \frac{\partial^{2} f}{\partial x^{2}} \cdot \frac{\partial^{2} f}{\partial y^{2}} - \frac{\partial^{2} f}{\partial x \partial y}$

The second derivatives of the function are:

$\frac{\partial^{2} f}{\partial x^{2}} = 0$

$\frac{\partial^{2} f}{\partial y^{2}} = 0$

$\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9$

Then, the expression is simplified to this and each point is tested:

$H = -16\cdot y +16\cdot x -9$

S1: (0,0)

$H = -9$ (Saddle Point)

S2: (3/8,-3/8)

$H = 3$ (Local maximum or minimum)

S3: (9/8, 0)

$H = 9$ (Local maximum or minimum)

S4: (0, - 9/8)

$H = 9$ (Local maximum or minimum)

Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

$f(\frac{3}{8} ,-\frac{3}{8} ) = - \frac{27}{64}$ (Local minimum)

S3: (9/8, 0)

$f(\frac{9}{8},0) = 0$ (Local maximum)

S4: (0, - 9/8)

$f(0,-\frac{9}{8} ) = 0$ (Local maximum)

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

4 0

3 years ago