Exercise 4.1. If the joint probability distribution of X and Y is given by r + y f(x, y) = 30 for r = 0, 1, 2, 3; y = 0, 1, 2, find (a) P(X < 2,Y = 1); (b) P(X > 2,Y < 1); (c) P(X > Y); (d) P(X +Y = 4).

Q: 1) Compute the centroid y bar with respect to its base.

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Q: Q3) In the following section: a) Calculate the coordinates of the center of gravity 8 6.

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Q: Problem 2: Determine the centroid X-bar with respect to Y-axis -150 mm 15 mm C 150 m -15 mm 15 mm…

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Q: y = kx³ I 30 mm Fig. P5.15 30 mm 150 mm X

A: Locate the centroid of the plane area shown.

Q: -3 R3 - 10 10 Find the coordinates of the centroid 3.

A: To determine the coordinates of the centroid of the given object.   The geometry of the given object…

Q: 150 mm Locate the centroid of the coloured area. 50 mm 100 mm 100 mm

A: Marking points over figure- A, B, C, D, E, F In the fig three shapes can be observed clearly- 1.…

Q: り Iy= ? 「ん

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Q: Determine the y-coordinate of the centroid of the shaded area shown. 9" Parabolic. 4" Answer: y = i…

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Q: 4x = y2 = 2 m

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Q: 44m 115 m 700 20m 100m 66m

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Q: A cube of edge 24mm is resting on the top of a frustum of cone of base diameter 60mm and top…

A: In the engineering graphics we can draw both the 2 dimensional projection and 3 dimensional…

Q: R=6 cm T. 5 cm 2 cm 8 cm (a)

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Q: Determine the location (x̄, y̅) of the centroid C of the area.

A: Consider the following diagrams for calculating the centroid C of the area:

Q: 4. Locate the centroid of the shaded area in the figure. y 3cm 6cm 4cm 8cm 9cm 4cm X

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Q: y 200 mm 300 mm → 125 mm 125 mm 60 mm 150 mm

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Q: Q.: Find the coordinate of the centriod of the shaded area 4 Y 3 m-

A: Given, Equation y=2x-30.5

Q: 1" 5" 2" 1-Determine the location of the centroid (x,y) of the centroid. 2- Determine the Ix0 and…

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Q: Determine the location of the centroidal x an y of the figure shown.

A: The portion which is cut out needs to be subtracted.

Q: Properties of moment of inertia of common shapes.

A: REFERENCE LINKS: https://engineeringstatics.org/MOI-common-shapes.html…

Q: -400 300 200, 200 Dimensions in mm 400 300 X

A: The given figure can be considered as (a) Rectangle, (b) Semi-circle and (c) Triangle

Q: Locate the centroid of the shaded area y r 3= 100 mm 50 mm 180 mm

A: Centroid is the point where we assume that whole mass of body is concentrated on single point.

Q: Let F₁ = [F1] N. i) Express F₁ as a Cartesian vector. F₁ X 3 5 4 30° 45° F₂= = 450 N - y

A: To represent F1 as a Cartesian vector. F1 = 647 N

Q: revolve the objects:   1) top view 2) right side view 3) front view

A: As per the guidelines we are supposed to answer only 3 subparts. Kindly repost the question for…

Q: Find the centroid of the area bounded by y2 = 4x, the y axis and y - 2 = 0. %3D

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Q: Find the x-coordinate of the centroid in 120 mm mm. - 60 mm - 40 mm 80 mm 60 mm

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Q: Find the centroid 20 30 25

A: A1=12×20×25=250 square unitA2=30×25=750 square unitA3=π×1522=353.43 square unit

Q: Locate the centroid for the following areas :

A: Figure is as shown

Q: B 75 mm 250 mm 40° A

A: Draw FBD: let us suppose AB as part 1 and BC as part 2: l1=250…

Q: Find the volume of excavation required for the area shown in the figure. Each grid point (20 in…

A: Depth of the cut at every location is given find the average depth of cut…

Q: 33.33 is wrong

A: The rigidities of walls A, B, and C  = 1.0 The rigidity of wall D = 1.5. Force acting in horizontal…

Q: Find the centroid of the given area. 5m 2m 3m 2m 8m

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Q: y | 12 kips 48 kip • ft D A B 4 ft- 4 ft 4 ft FIG. P4.16

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Q: the coordinates of the centroid of the shaded areas

A: To find the coordinates of the centroids

Q: -30- 20 60 30 60 - 60 60 Dimensions in mm

A: you can clearly understand if you follow the below solution

Q: Solve for x 5-5x>7x+16

A: Given: 5-5x&gt;7x+16

Types of Composite

Composite materials are formed by merging two components that have different chemical and physical properties. The merged output is termed as a composite, whose properties are completely different from the individual components. Composite materials also have better strength than the individual components.Composite materials usually have two main parts, reinforcement material, and matrix material. The matrix material acts as a binder and surrounds the fiber of another material that bears a much higher strength. This material is known as the reinforcement material. The matrix materials are usually high-strength additives resins, while the fibers of the reinforced material can be random, oriented, and in a mat format.Composite materials are also known as fiber-reinforced composite polymers because they are composed of the polymer matrix which is reinforced to laboratory-created fibers or naturally occurring fibers, such as glass fibers, carbon fibers, or aramid fibers. Under the application of external loads, the matrix material absorbs and transfers the loads to the fibers, these fibers in return provide good compressive strength, tensile strength, and stiffness to the fiber-reinforced composite materials.Depending upon the matrix and the reinforcement material, different properties can be induced to the fiber-reinforced composite materials, including good corrosion resistance properties. The most common matrix material is the resin while the glass fibers act as the reinforcement material. Strong fibers provide strength and stiffness to the composite materials while, a soft and flexible resin provides shape to the composite materials.