What is the answer I need ASAP

What is the solution for x in the inequality 4−2x>8

ElenaW [278]

Answer:

X<-2

Step-by-step explanation:

7 0

3 years ago

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You’re giving a side length of 7 cm. How many equilateral triangle’s can you construct using this information

dalvyx [7]

Answer:

1

Step-by-step explanation:

You can't construct more than 1 triangle. If either of the angles shift, the triangle won't close. And a equilateral triangle must close to be considered, well, a triangle.

3 0

3 years ago

A person purchased a slot machine and tested it by playing it 1,137 times. There are 10 different categories of outcomes, includ

ivann1987 [24]

Answer:

$p_v= P(\chi^2_{9}>11.517)=0.2419$

And on this case if we see the significance level given $\alpha=0.1$ we see that $p_v>alpha$ so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

Step-by-step explanation:

A chi-square goodness of fit test determines if a sample data obtained fit to a specified population.

$p_v$ represent the p value for the test

O= obserbed values

E= expected values

The system of hypothesis for this case are:

Null hypothesis: $O_i = E_i[/tex[Alternative hypothesis: [tex]O_i \neq E_i$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

On this case after calculate the statistic they got: $\chi^2 = 11.517$

And in order to calculate the p value we need to find first the degrees of freedom given by:

$df=n-1=10-1=9$ , where k represent the number of levels (on this cas we have 10 categories)

And in order to calculate the p value we need to calculate the following probability:

$p_v= P(\chi^2_{9}>11.517)=0.2419$

And on this case if we see the significance level given $\alpha=0.1$ we see that $p_v>alpha$ so we fail to reject the null hypothesis that the observed outcomes agree with the expected frequencies at 10% of significance.

6 0

3 years ago

A 63 foot long cable is anchored in the ground and

myrzilka [38]

Answer:

The height of the light pole to the nearest foot is approximately 50 feet

Step-by-step explanation:

The parameters given in the question are;

The length of the cable = 63 feet

The angle of elevation of the cable to the top of the light pole = 52°

With the assumption that the light pole is perpendicular to the ground, the figure formed by the light pole, the distance of the of the cable from the base of the light pole and the length of the cable form a right triangle

The question can be answered by using trigonometric relations for the sine of the given angle as follows;

$sin ( \theta) = \dfrac{Opposite \ leg \ length}{Hypotenuse \ length}$

The opposite leg length of the formed right triangle = The height of the light pole

The hypotenuse length = The length of the cable = 63 feet

The angle, θ = The angle of elevation = 52°

Plugging in the values, gives;

$sin (52 ^ {\circ}) = \dfrac{The \ height \ of \ the \ light \ pole }{63 \ feet}$

∴ The height of the light pole = 63 feet × sin(52°) ≈ 49.645 feet

The height of the light pole to the nearest foot ≈ 50 feet.

8 0

3 years ago

Video Example EXAMPLE 3 Find the local maximum and minimum values and saddle points of f(x, y) = x4 + y4 − 4xy + 1. SOLUTION We

forsale [732]

Answer:

(0, 0) is a saddle point

(1, 1) is a local minimum

(-1, -1) is another point of local minimum

Step-by-step explanation:

We first locate the critical points. In order to get the critical points we need to find the first derivatives and then set them to zero.

f(x, y) = x⁴ + y⁴ - 4 xy + 1

Find the first derivatives wrt to x and y

$f_{x}(x,y)$ = 4x ³ - 4y --> (1)

$f_{y}(x,y)$ = 4y³ - 4x --> (2)

Solve (1) for y

4x ³ - 4y = 0

x ³ = y

y = x ³ ---> (3)

Solve (2) for x

4y³ - 4x = 0

4y³ = 4x

y³ = x

x = y³ ----> (4)

Plug (3) into (2)

4y³ - 4x

4(x ³)³ - 4x = 0

4x⁹ - 4x = 0

4x (x ⁸ - 1) = 0

4x (x ⁴ - 1)(x⁴ + 1) = 0

4x (x ² - 1)(x ² + 1)(x ⁴ + 1) = 0

4x (x - 1)(x + 1)(x ² + 1)(x ⁴ + 1) = 0

So

x = 1 , -1 , 0

In order to find values of y, Plug each x value in (3)

For

x = 0

corresponding y value

y = x ³

y = 0 ³

y = 0

For

x = 1

y = x ³

y = 1 ³

y = 1

For

x = -1

y = x ³

y = (-1 )³

y = -1

Hence we get the critical points which are:

(0, 0)

(1, 1) and

(-1, -1)

Now for each critical point, we have to compute D(x,y)

For a critical point (x,y), D computed as:

D(x, y) = $f_{xx}$ (x, y) - $f_{yy}$ (x, y) - ( $f_{xy}$ (x, y))²

After computing D(x,y) check:

If D(x, y) > 0 and $f_{xx}$ (x, y) > 0:

f(x, y) is a local minimum.

If D(x, y) > 0 and $f_{xx}$ (x, y) < 0:

f(x, y) is a local maximum.

If D(x, y) < 0:

then f(x, y) is a saddle point

Here first compute the second derivative in order to get $f_{xx}$ , $f_{yy}$ and $f_{xy}$

we have already computed:

$f_{x}(x,y)$ = 4x ³ - 4y --> (1)

$f_{y}(x,y)$ = 4y³ - 4x --> (2)

Now

$f_{xx}(x,y)$ = 12x²

$f_{yy}(x,y)$ = 12y²

$f_{xy}(x,y)$ = -4

Compute D(x,y)

Critical Point (0, 0):

$D(0, 0) = f_{xx} (0,0) f_{yy}(0,0)-(f_{xy}(0,0))^{2}$

Putting values of $f_{xx}(x,y)$ , $f_{yy}(x,y)$ and $f_{xy}(x,y)$ in above equation:

D(0,0) = 12(0)² * 12(0)² - (-4)²

= 0 * 0 -16

D(0,0) = -16

We know that if D < 0, the critical point f(x, y) is a saddle point.

D(0,0) < 0 because D(0,0) = -16

Hence (0, 0) is a saddle point

Compute D(x,y)

Critical Point (1, 1):

$D(1, 1) = f_{xx} (1,1) f_{yy}(1,1)-(f_{xy}(1,1))^{2}$

Putting values of $f_{xx}(x,y)$ , $f_{yy}(x,y)$ and $f_{xy}(x,y)$ in above equation:

D(1,1) = 12(1)² * 12(1)² - (-4)²

= 12 * 12 - 16

D(1,1) = 128

We know that if D(x, y) > 0 and $f_{xx}$ (x, y) > 0 then f(x, y) is a local minimum.

D(1,1) > 0 because D(1,1) = 128

$f_{xx}$ (x, y) > 0 because $f_{xx}$ (x, y) = 12

Hence (1,1) is the local minimum

Compute D(x,y)

Critical Point (-1, -1):

$D(-1, -1) = f_{xx} (-1,-1) f_{yy}(-1,-1)-(f_{xy}(-1,-1))^{2}$

Putting values of $f_{xx}(x,y)$ , $f_{yy}(x,y)$ and $f_{xy}(x,y)$ in above equation:

D(-1,-1) = 12(-1)² * 12(-1)² - (-4)²

= 12 * 12 - 16

D(-1,-1) = 128

We know that if D(x, y) > 0 and $f_{xx}$ (x, y) > 0 then f(x, y) is a local minimum.

D(-1,-1) > 0 because D(-1,-1) = 128

$f_{xx}$ (x, y) > 0 because $f_{xx}$ (x, y) = 12

Hence (-1,-1) is the local minimum

6 0

3 years ago