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3 years ago

Can someone please help me g(x) = 17 -*; g(6)

Mathematics

2 answers:

Sav [38]3 years ago

6 0

Answer:

Divide each term in G(6)=17−(2/3)x by 6.

G(6)/6=17/6−(2/3)x⋅16

Cancel the common factor of 6.

G=17/6−(2/3)x⋅16

Simplify each term.

G=17/6−x/9

Step-by-step explanation:

DerKrebs [107]3 years ago

3 0

Answer:

(x + 6)(x-4) = -16

x² + 2x -24 = -16

x² + 2x - 8 = 0

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

x = - 4 or x = 2

Step-by-step explanation:

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Point A is inside triangle E D F. Lines are drawn from the points of the triangle to point A. Lines are drawn from point A to th

Crazy boy [7]

Answer:

Option B. 3 is the answer.

Step-by-step explanation:

To draw a circumcenter of a triangle we draw the perpendicular bisectors of the the sides of the triangle first.

Intersection point of these perpendiculars drawn is the circumcenter of the circle.

Distances of the circumcenter to the vertices are same, that means distance of circumcenter from the vertex is the radius of the circle.

Therefore, length of AE and AD will be same in measure.

AE = AD

2n + 4 = 10

2n = 10 - 4

2n = 6

n = 3

Therefore, Option B. 3 is the answer.

5 0

3 years ago

Based on the number of voids, a ferrite slab is classified as either high, medium, or low. Historically, 5% of the slabs are cla

AnnyKZ [126]

Answer:

(a) Name: Multinomial distribution

Parameters: $p_1 = 5\%$ $p_2 = 85\%$ $p_3 = 10\%$ $n = 20$

(b) Range: $\{(x,y,z)| x + y + z=20\}$

Parameters: $p_1 = 5\%$ $n = 20$

$(d)\ E(x) = 1$ $Var(x) = 0.95$

$(e)\ P(X = 1, Y = 17, Z = 3) = 0$

$(f)\ P(X \le 1, Y = 17, Z = 3) =0.07195$

$(g)\ P(X \le 1) = 0.7359$

$(h)\ E(Y) = 17$

Step-by-step explanation:

Given

$p_1 = 5\%$

$p_2 = 85\%$

$p_3 = 10\%$

$n = 20$

$X \to$ High Slabs

$Y \to$ Medium Slabs

$Z \to$ Low Slabs

Solving (a): Names and values of joint pdf of X, Y and Z

Given that:

$X \to$ Number of voids considered as high slabs

$Y \to$ Number of voids considered as medium slabs

$Z \to$ Number of voids considered as low slabs

Since the variables are more than 2 (2 means binomial), then the name is multinomial distribution

The parameters are:

$p_1 = 5\%$ $p_2 = 85\%$ $p_3 = 10\%$ $n = 20$

And the mass function is:

$f_{XYZ} = P(X = x; Y = y; Z = z) = \frac{n!}{x!y!z!} * p_1^xp_2^yp_3^z$

Solving (b): The range of the joint pdf of X, Y and Z

Given that:

$n = 20$

The number of voids (x, y and z) cannot be negative and they must be integers; So:

$x + y + z = n$

$x + y + z = 20$

Hence, the range is:

$\{(x,y,z)| x + y + z=20\}$

Solving (c): Names and values of marginal pdf of X

We have the following parameters attributed to X:

$p_1 = 5\%$ and $n = 20$

Hence, the name is: Binomial distribution

Solving (d): E(x) and Var(x)

In (c), we have:

$p_1 = 5\%$ and $n = 20$

$E(x) = p_1* n$

$E(x) = 5\% * 20$

$E(x) = 1$

$Var(x) = E(x) * (1 - p_1)$

$Var(x) = 1 * (1 - 5\%)$

$Var(x) = 1 * 0.95$

$Var(x) = 0.95$

$(e)\ P(X = 1, Y = 17, Z = 3)$

In (b), we have: $x + y + z = 20$

However, the given values of x in this question implies that:

$x + y + z = 1 + 17 + 3$

$x + y + z = 21$

Hence:

$P(X = 1, Y = 17, Z = 3) = 0$

$(f)\ P{X \le 1, Y = 17, Z = 3)$

This question implies that:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3) + P(X = 1, Y = 17, Z = 3)$

Because

$0, 1 \le 1$ --- for x

In (e), we have:

$P(X = 1, Y = 17, Z = 3) = 0$

So:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3) +0$

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3)$

In (a), we have:

$f_{XYZ} = P(X = x; Y = y; Z = z) = \frac{n!}{x!y!z!} * p_1^xp_2^yp_3^z$

So:

$P(X=0; Y=17; Z = 3) = \frac{20!}{0! * 17! * 3!} * (5\%)^0 * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20!}{1 * 17! * 3!} * 1 * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20!}{17! * 3!} * (85\%)^{17} * (10\%)^{3}$

Expand

$P(X=0; Y=17; Z = 3) = \frac{20*19*18*17!}{17! * 3*2*1} * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = \frac{20*19*18}{6} * (85\%)^{17} * (10\%)^{3}$

$P(X=0; Y=17; Z = 3) = 20*19*3 * (85\%)^{17} * (10\%)^{3}$

Using a calculator, we have:

$P(X=0; Y=17; Z = 3) = 0.07195$

So:

$P(X \le 1, Y = 17, Z = 3) =P(X = 0, Y = 17, Z = 3)$

$P(X \le 1, Y = 17, Z = 3) =0.07195$

$(g)\ P(X \le 1)$

This implies that:

$P(X \le 1) = P(X = 0) + P(X = 1)$

In (c), we established that X is a binomial distribution with the following parameters:

$p_1 = 5\%$ $n = 20$

Such that:

$P(X=x) = ^nC_x * p_1^x * (1 - p_1)^{n - x}$

So:

$P(X=0) = ^{20}C_0 * (5\%)^0 * (1 - 5\%)^{20 - 0}$

$P(X=0) = ^{20}C_0 * 1 * (1 - 5\%)^{20}$

$P(X=0) = 1 * 1 * (95\%)^{20}$

$P(X=0) = 0.3585$

$P(X=1) = ^{20}C_1 * (5\%)^1 * (1 - 5\%)^{20 - 1}$

$P(X=1) = 20 * (5\%)* (1 - 5\%)^{19}$

$P(X=1) = 0.3774$

So:

$P(X \le 1) = P(X = 0) + P(X = 1)$

$P(X \le 1) = 0.3585 + 0.3774$

$P(X \le 1) = 0.7359$

$(h)\ E(Y)$

Y has the following parameters

$p_2 = 85\%$ and $n = 20$

$E(Y) = p_2 * n$

$E(Y) = 85\% * 20$

$E(Y) = 17$

8 0

3 years ago

Factor completely 9r^3-72r^2

PtichkaEL [24]

Answer:

9r²(r - 8)

Step-by-step explanation:

Step 1: Write expression

9r³ - 72r²

Step 2: Factor out 9

9(r³ - 8r²)

Step 3: Factor out r²

9r²(r - 8)

6 0

3 years ago

.........................................

Veseljchak [2.6K]

Answer:

Whats wrong??

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Find the solutions of the equation x4 − 2x3 − x2 + 2x = 0. (Enter your answers as a comma-separated list.)

Bad White [126]

Answer:

$S=\left \{ 0,-1,1,2 \right \}$

Step-by-step explanation:

There are several ways to solve this quartic equation. But since the coefficients, they repeat a=1,b=2,c=1,d=2, but e=0, and they are multiple of each other, then it is more convenient to work with factoring as the method of solving it.

As if it was a quadratic one.

$x^{4}-2x^{3}- x^{2} + 2x = 0\\x(x^{3}-2x^{2}-x+2)=0 \:Factoring \:out\\x[(x^{3}-2x^{2})+(-x+2)]=0 \:Grouping\\x[\mathbf{x^{2}}(x-2)+\mathbf{-1}(x+2)]=0 \:Rewriting\:the\:first\:factor\\x(x^{2}-1)(x-2)\:Expanding \:the \:first \:factor\\x(x-1)(x+1)(x-2)=0\\x=0,x=1,x=-1,x=2\\S=\left \{ 0,-1,1,2 \right \}$

7 0

3 years ago

Read 2 more answers