Evaluate : (-2)^5 × (-2)^-8 ( do as director )

The animal shelter has both male and female Labrador retrievers in yellow, brown, or black. There is and equal number of each ki

Illusion [34]

Answer:

Step-by-step explanation:

There are choices when it comes to color (3). Then there is the choice between a boy dog or a girl dog (2). Multiply the number of colors by the number of genders. (6)

When choosing a dog from random, you can either choose:

Brown/M

Brown/F

Black/M

Black/F

Yellow/M

Yellow/F

(This list is called the Sample Space, a list of all possible outcomes)

Because the text says that all types of dogs are found in even numbers, it would not matter if there was 6, 12, 18, ... amount of dogs.

When looking at this sample space, you can see that the chance of choosing a Yellow/F dog, is 1 in 6.

5 0

3 years ago

Write and solve a real-word word problem in which there are 3 tables for every 8 chairs

Talja [164]

Answer:

tate is hosting a party and he has 23 people coming over. for every table he buys, 8 chairs come with it. how many tables should tate buy so that he can seat all of his guests?

Step-by-step explanation:

6 0

3 years ago

Rolling 7 or 11 on the first roll is a win probability of winning on 1st roll

gtnhenbr [62]

1/6 plus 1/18, which equals to 2/9. Suppose we roll 4 on the first roll (the probability of rolling 4 is 1/12). On each successive roll the probability of rolling 7 is 1/6 and the probability of rolling 4 is 1/12.

[i.dk how this can help but if you got a 7 or 11 rolling on the first round you would probably win]

5 0

3 years ago

Somebody please help me

Mazyrski [523]

Plan A: 30 + 0.01m
Plan B: 20 + 0.05m
in order for 2 plans cost the same, both need to be equal
so should be

20 + 0.05m = 30 + 0.01m

answer is D. last one
20 + 0.05m = 30 + 0.01m

8 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

Evaluate : (-2)^5 × (-2)^-8 ( do as director )​

Evaluate : (-2)^5 × (-2)^-8 ( do as director )