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3 years ago

40 points, please help with all of them

Mathematics

1 answer:

Kitty [74]3 years ago

6 0

Answer:

$\large\boxed{Q1.\ \text{2 times greater}}\\\\\boxed{Q2.\ J. 3}\\\\\boxed{Q3.\ B. -\dfrac{4}{5}}$

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

m - slope

b - y-intercept

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

Q1.

We have the equation of a line p in the standard form

$4x-3y=7$

Convert to the slope-intercept form:

$4x-3y=7$ subtract 3x from both sides

$-3y=-4x+7$ divide both sides by (-3)

$y=\dfrac{4}{3}x-\dfrac{7}{3}$

The slope $m_1=\dfrac{4}{3}$

From the table we have the points (4, 3) and (7, 5). Calculate the slope of line q:

$m_2=\dfrac{5-3}{7-4}=\dfrac{2}{3}$

Divide the slope of p by the slope of q:

$\dfrac{4}{3}:\dfrac{2}{3}=\dfrac{4}{3}\cdot\dfrac{3}{2}=2$

Q2.

Parallel line have the same slope. Therefore, if we have the equation of the line in the slope-intercept form, then we have the slope:

$y=3x+2\to m=3$

Q3.

Parallel line have the same slope.

Calculate the slope from given points (-11, 5) and (-6, 1):

$y=\dfrac{1-5}{-6-(-11)}=\dfrac{-4}{-6+11}=\dfrac{-4}{5}=-\dfrac{4}{5}$

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Please help me !!!!!! I need help

kvv77 [185]

Answer:

y +8 = -4(x -8)

Step-by-step explanation:

You recognize that the given equation is in slope-intercept form:

y = mx + b

with m = 1/4 and b = 5.

A perpendicular line will have a slope that is the negative reciprocal of this value of m, so the desired slope is ...

-1/m = -1/(1/4) = -4

The point-slope form of the equation for a line is ...

y -k = m(x -h) . . . . . for slope m through point (h, k)

Using m=-4 and (h, k) = (8, -8), the point-slope form of the equation for the line you want is ...

y +8 = -4(x -8)

3 0

2 years ago

Analyze the graph h(x).

Dominik [7]

Answer:

D h(x) = f(x)×g(x)

Step-by-step explanation:

h(x) has a wave with 2 changes in direction.

so, this needs to be an expression of the third degree (there must be a term with x³ as the highest power of x).

and that is only possible when multiplying both basic functions. all the other options would keep it at second degree (x²) or render it even to a first degree (linear).

8 0

2 years ago

Kindly answer this one. Thank you.

steposvetlana [31]

Answer:

See below.

Step-by-step explanation:

I'm assuming these questions are about the Midline Theorem (segment AL joins the midpoints of the non-parallel sides.

♦ The midline's length is the average of the lengths of the top and bottom parallel sides.

$AL=\frac{OR+CE}{2}$

Use this equation and substitute values given in each problem, then solve for the missing information.

1. AL = x, CE = 9, OR = 5

$x=\frac{9+5}{2}=7$

2. AL = m - 4, CE = 15, OR = 17

$m-4=\frac{15+17}{2}=16\\\\m-4=16\\\\\\m=20\\\\AL=20-4=16$

3. OR = y + 5, AL = 15, CE = 18

$15=\frac{(y+5)+18}{2}\\\\15=\frac{y+23}{2}\\\\30=y+23\\\\7=y\\\\OR = 7+5=12$

4 0

2 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

2 years ago

A swimming pool is 30 ft long and 15 ft wide. How much water do you need to fill the pool if it is 6 ft deep? (Thanks?)

VMariaS [17]

Well the volume of the pool is 2700 ft

3 0

2 years ago