A rock is dropped from a height of 100 feet. Calculate the time between when the rock was dropped and when it landed. If we choose "down" as positive and ignore air friction, the function is h(t) = 25t^2-81

Answer:

tanA = [tex]\frac{55}{48}[/tex]

Step-by-step explanation:

tanA = [tex]\frac{opposite}{adjacent}[/tex] = [tex]\frac{BC}{AC}[/tex] = [tex]\frac{55}{48}[/tex]

Answer:

B' (-6, 7)

C' ( -8, 1)

Step-by-step explanation:

The rule is

(x,y) → (x -2, y - 1)

A( -2,4) → A' ( -4,3)

To get from A to A', the x value changed by -2 (-2-2 = -4).  The y changed by -1 ( 3-1 = 3)

Helping in the name of Jesus.

The tangential component of the acceleration vector is (4t / (1 + 4t² + 9)[tex]^{1/2}[/tex])i + (8t²/ (1 + 4t² + 9)[tex]^{1/2}[/tex])j + (12t / (1 + 4t² + 9)[tex]^{1/2}[/tex])k, and the normal component of the acceleration vector is -4t / (1 + 4t² + 9)[tex]^{1/2}[/tex] * i + (2 - 8t² / (1 + 4t² + 9)[tex]^{1/2}[/tex])j - 12t / (1 + 4t² + 9)[tex]^{1/2}[/tex] * k.

To find the tangential and normal components of the acceleration vector, we first need to find the acceleration vector itself by taking the second derivative of the position vector r(t):

r(t) = ti + [tex]t^{2j}[/tex] + 3tk

v(t) = dr/dt = i + 2tj + 3k

a(t) = dv/dt = 2j

The acceleration vector is a(t) = 2j. This means that the acceleration is entirely in the y-direction, and there is no acceleration in the x- or z-directions.

The tangential component of the acceleration vector, a_T, is the component of the acceleration vector that is parallel to the velocity vector v(t). Since the velocity vector is i + 2tj + 3k and the acceleration vector is 2j, the tangential component is:

a_T = (a(t) · v(t) / ||v(t)||[tex]^{2}[/tex]) * v(t) = (0 + 4t + 0) / [tex](1 + 4t^{2} + 9)^{1/2}[/tex] * (i + 2tj + 3k)

Simplifying this expression, we get:

a_T = (4t / [tex](1 + 4t^{2} + 9 ^{1/2} )[/tex]i + (8t^2 / (1 + 4t^2 + 9)^(1/2))j + (12t / (1 + 4t^2 + 9)[tex]^{1/2}[/tex])k

The normal component of the acceleration vector, a_N, is the component of the acceleration vector that is perpendicular to the velocity vector. Since the acceleration vector is entirely in the y-direction, the normal component is:

a_N = a(t) - a_T = -4t / (1 + 4t² + 9)[tex]^{1/2}[/tex]* i + (2 - 8t² / (1 + 4t²+ 9)[tex]^{1/2}[/tex])j - 12t / (1 + 4t² + 9)[tex]^{1/2}[/tex]* k

Therefore, the tangential component of the acceleration vector is (4t / (1 + 4t² + 9)[tex]^{1/2}[/tex])i + (8t²/ (1 + 4t² + 9)[tex]^{1/2}[/tex])j + (12t / (1 + 4t² + 9)[tex]^{1/2}[/tex])k, and the normal component of the acceleration vector is -4t / (1 + 4t² + 9)[tex]^{1/2}[/tex] * i + (2 - 8t² / (1 + 4t² + 9)[tex]^{1/2}[/tex])j - 12t / (1 + 4t² + 9)[tex]^{1/2}[/tex] * k.

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The exponential function that models the population of Trenton, NJ since 1980 is: P(t) = 92124 * [tex](1-0.0272)^t[/tex]

1. The initial population in 1980 is given as 92,124.

2. The rate of decrease is 2.72% or 0.0272 in decimal form.

3. Since the population is decreasing, we subtract the rate from 1 (1 - 0.0272 = 0.9728).

4. The exponential function is written in the form P(t) = P₀ * [tex](1 +r)^t[/tex] , where P₀ is the initial population, r is the rate of change, and t is the number of years after 1980.

5. In this case, P₀ = 92124, r = -0.0272, and we want to find the population at time t.

6. Therefore, the exponential function that models this situation is P(t) = 92124 * [tex](0.9728)^t[/tex] .

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Answer:

c. 50(i-0.1)² + 210(i - 0.1) + 320.5.

Step-by-step explanation:

To find the equivalent expression that contains the future value for a monthly interest rate of i = 0.1, we simply substitute i = 0.1 into the equation V = 50(1 + i)² + 100(1 + i) + 150 and simplify.

V = 50(1 + 0.1)² + 100(1 + 0.1) + 150

V = 50(1.1)² + 100(1.1) + 150

V = 50(1.21) + 110 + 150

V = 60.5 + 110 + 150

V = 320.5

Therefore, the expression that contains the future value for a monthly interest rate of i = 0.1 is c. 50(i-0.1)² + 210(i - 0.1) + 320.5.

Based on the information provided, the hydrocarbon that fits these criteria is likely to be octane, with a molecular formula of C8H18. The molecular ion with an m/z value of 128 indicates that the molecule has lost one electron, resulting in a positive charge.

The base peak with an m/z value of 43 is likely due to the fragmentation of a methyl group (CH3) from the parent molecule. The significant peaks with m/z values of 57, 71, and 85 may correspond to other fragment ions resulting from the breakdown of the octane molecule.
Based on the given m/z values, the hydrocarbon you are looking for has a molecular ion with an m/z value of 128, a base peak with an m/z value of 43, and significant peaks with m/z values of 57, 71, and 85. The hydrocarbon is likely an alkane, alkene, or alkyne. To determine the exact compound, further information such as the chemical formula or structure would be needed.

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The steady state vector ¯ q for the given stochastic matrix p such that p ¯ q = ¯ q is q1 = 3q2, where q2 can be any real number.

The steady state vector, denoted as ¯ q, for the given stochastic matrix p, such that p ¯ q = ¯ q, can be found by solving for the eigenvector corresponding to the eigenvalue of 1 for matrix p.

Start with the given stochastic matrix p:

p = [ 0.9 0.3 ]

[ 0.1 0.7 ]

Next, subtract the identity matrix I from p, where I is a 2x2 identity matrix:

p - I = [ 0.9 - 1 0.3 ]

[ 0.1 0.7 - 1 ]

Find the eigenvalues of (p - I) by solving the characteristic equation det(p - I) = 0:

| 0.9 - 1 0.3 | | -0.1 0.3 | | -0.1 * (0.7 - 1) - 0.3 * 0.1 | | -0.1 - 0.03 | | -0.13 |

| 0.1 0.7 - 1 | = | 0.1 -0.3 | = | 0.1 * 0.1 - (0.7 - 1) * 0.3 | = | 0.1 + 0.27 | = | 0.37 |

Therefore, the eigenvalues of (p - I) are -0.13 and 0.37.

Solve for the eigenvector corresponding to the eigenvalue of 1. Substitute λ = 1 into (p - I) ¯ q = 0:

(p - I) ¯ q = [ -0.1 0.3 ] [ q1 ] = [ 0 ]

[ 0.1 -0.3 ] [ q2 ] [ 0 ]

This results in the following system of linear equations:

-0.1q1 + 0.3q2 = 0

0.1q1 - 0.3q2 = 0

Solve the system of linear equations to obtain the eigenvector ¯ q:

By substituting q1 = 3q2 into the first equation, we get:

-0.1(3q2) + 0.3q2 = 0

-0.3q2 + 0.3q2 = 0

0 = 0

This shows that the system of equations is dependent and has infinitely many solutions. We can choose any value for q2 and calculate the corresponding q1 using q1 = 3q2.

Therefore, the steady state vector ¯ q is given by:

q1 = 3q2

q2 = any real number

In conclusion, the steady state vector ¯ q for the given stochastic matrix p such that p ¯ q = ¯ q is q1 = 3q2, where q2 can be any real number.

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If the "swimming-pool" for children is built with rectangular-prism and 2 halves of cylinder, then the total volume of pool is 312.64 m³.

From the figure, we observe that the swimming pool is made up of a rectangular prism, and 2 halves of cylinder,

the diameter of the half of cylinder is = 16m ,

So, radius of the half of cylinder is = 16/2 = 8m,

The volume of 2 halves of cylinder is = πr²h,

Substituting the values,

We get,

Volume of 2 halves of cylinder is = π × 8 × 8 × 0.6 ≈ 120.64 m³,

Now, the volume of the rectangular prism is = 20 × 16 × 0.6 = 192 m³,

So, the Volume of the swimming pool is = 192 + 120.64 = 312.64 m³.

Therefore, the total volume of swimming pool is 312.64 m³.

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The closest linear combination of (1, 2, -1) and (1, 0, 1) to b is:

(3/4, 0, 3/4)

To find the linear combination of (1, 2, -1) and (1, 0, 1) that is closest to b = (2, 1, 1), we can use the projection formula:

proj_u(b) = ((b . u) / (u . u)) * u

where u is one of the vectors we are using to form the linear combination, and "." denotes the dot product.

Let's start by finding the projection of b onto (1, 2, -1):

proj_(1,2,-1)(2,1,1) = ((2,1,1) . (1,2,-1)) / ((1,2,-1) . (1,2,-1)) * (1,2,-1)

= (0) / (6) * (1,2,-1)

= (0,0,0)

Since the projection of b onto (1, 2, -1) is the zero vector, we know that (1, 2, -1) is orthogonal to b. This means that the closest linear combination of (1, 2, -1) and (1, 0, 1) to b will only involve (1, 0, 1).

Let's find the projection of b onto (1, 0, 1):

proj_(1,0,1)(2,1,1) = ((2,1,1) . (1,0,1)) / ((1,0,1) . (1,0,1)) * (1,0,1)

= (3/2) / (2) * (1,0,1)

= (3/4,0,3/4)

So the closest linear combination of (1, 2, -1) and (1, 0, 1) to b is:

(3/4, 0, 3/4)

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The problem involves finding the compositions of two functions f and g, their inverse functions, and the composition of the inverse functions. The solution demonstrates how to apply these concepts.

To find the compositions of functions and their inverse functions.

Using the given definitions of f and g.

We find their compositions and their inverse functions. Then we apply these results to find the compositions of inverse functions.

(a) fog: [tex]fog(x) = f(g(x)) = f(5x+7) = 2(5x+7) + 3 = 10x + 17[/tex]

(b) gof: [tex]gof(x) = g(f(x)) = g(2x+3) = 5(2x+3) + 7 = 10x + 22[/tex]

(c) [tex](fog)^-1:[/tex]

We first find fog(x) and then solve for x: [tex]fog(x) = 10x + 17[/tex]

[tex]x = (fog(x) - 17)/10[/tex]

[tex](fog)^-1(x) = (x - 17)/10[/tex]

[tex](d) f^-1 o g^-1:[/tex]

[tex]f^-1(x) = (x - 3)/2[/tex]

[tex]g^-1(x) = (x - 7)/5[/tex]

[tex](f^-1 o g^-1)(x) = f^-1(g^-1(x)) = f^-1((x-7)/5)[/tex] = [tex][(x-7)/5 - 3]/2 = (x-23)/10[/tex]

(e)[tex]g^1 o f^-1:[/tex] [tex]g^1(x) = (x-7)/5[/tex]

[tex](g^1 o f^-1)(x) = g^1(f^-1(x))[/tex]

=[tex]g^1((x-3)/2) = 5((x-3)/2) + 7[/tex]

=[tex](5x+2)/2[/tex]

Overall, the problem requires a solid understanding of function composition, inverse functions, and basic algebraic manipulation.

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a) Cylindrical coordinates for point (-5, 5, 5) are (r, θ, z) = (√50, 3π/4, 5).
b) Cylindrical coordinates for point (-3, 3√3, 1) are (r, θ, z) = (6, 5π/6, 1).

We will use the following equations:

1. r = √(x² + y²)
2. θ = arctan(y/x) (note: make sure to take the quadrant into account)
3. z = z (z-coordinate remains the same)

(a) For the point (-5, 5, 5):

1. r = √((-5)² + 5²) = √(25 + 25) = √50
2. θ = arctan(5/-5) = arctan(-1) = 3π/4 (in the 2nd quadrant)
3. z = 5

So, the cylindrical coordinates for point (-5, 5, 5) are (r, θ, z) = (√50, 3π/4, 5).

(b) For the point (-3, 3√3, 1):

1. r = √((-3)² + (3√3)²) = √(9 + 27) = √36 = 6
2. θ = arctan((3√3)/-3) = arctan(-√3) = 5π/6 (in the 2nd quadrant)
3. z = 1

So, the cylindrical coordinates for point (-3, 3√3, 1) are (r, θ, z) = (6, 5π/6, 1).

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The number which a value, 20 percent less than 120 is one-third more than is 72

A percentage is an expression of the ratio between quantities, expressed as a fraction with a denominator of 100.

The quantity 20 percent less than 120 can be expressed as follows;

20 percent less than 120 = ((100 - 20)/100) × 120 = 96

One-third more than a number = The number + (The number)/3

Let x represent the number, we get;

One-third more than the number = x + x/3

x + x/3 = 96

x·(1 + 1/3) = 96

4·x/3 = 96

x = 96 × 3/4 = 72

Therefore, 20 percent less than 120 is one-third more than 72

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The cost of the film for the whole area of the figure is $73.6.

Given that,

Chris is covering a window with a decorative adhesive film to filter light.

The figure is a window in the shape of a parallelogram.

We have to find the area of the figure.

Area of parallelogram = Base × Height

Area = 8 × 4 = 32 feet²

Cost for the film per square foot = $2.3

Cost of the film for 32 square foot = 32 × $2.3 = $73.6

Hence the cost of the film is $73.6.

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The function with an amplitude of 3 and a phase shift of π/2 is h(x) = 3 cos (2x - π/2) + 3.

The amplitude of a function is the distance between the maximum and minimum values of the function, divided by 2. The phase shift of a function is the horizontal shift of the function from the standard position,

(y = cos(x) or y = sin(x)).
To find the function with an amplitude of 3 and a phase shift of π/2, we need to look for a function that has a coefficient of 3 on the cosine term and a horizontal shift of π/2.
Looking at the given options, we can eliminate option a) because it has a coefficient of -3 on the cosine term, which means that its amplitude is 3 but it is inverted.

Option b) has a coefficient of 3 on the cosine term but it has a phase shift of -π/2, which means it is shifted to the left instead of to the right. Option d) has a phase shift of π/2, but it has a coefficient of -2 on the cosine term, which means its amplitude is 2 and not 3.
A*cos(B( x - C)) + D

Where A is the amplitude, B affects the period, C is the phase shift, and D is the vertical shift.

f(x) = -3 cos(2x - π) + 4

Amplitude: |-3| = 3

Phase shift: π (not π/2) b) g(x) = 3cos(2x + π) -1

Amplitude: |3| = 3

Phase shift: -π (not π/2) c) h(x) = 3 cos (2x - π/2) + 3

Amplitude: |3| = 3

Phase shift: π/2 d) j(x) = -2cos(2x + π/2) + 3200

Amplitude: |-2| = 2 (not 3)

Phase shift: -π/2
Therefore, the only option left is c) h(x) = 3 cos (2x - π/2) + 3. This function has a coefficient of 3 on the cosine term and a horizontal shift of π/2, which means it has an amplitude of 3 and a phase shift of π/2.
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The shortest distance from the point (3, 0, −2) to the plane x + y + z = 2 is √(3) or approximately 1.732 units.

To find the shortest distance, d, from the point (3, 0, −2) to the plane x + y + z = 2, we need to use the formula for the distance between a point and a plane.

First, we need to find the normal vector of the plane. The coefficients of x, y, and z in the plane equation (1, 1, 1) form the normal vector (since the plane is perpendicular to this vector).

Next, we can use the point-to-plane distance formula:

d = |(ax + by + cz - d) / √(a² + b² + c²)|

where (a, b, c) is the normal vector of the plane, (x, y, z) is the coordinates of the point, and d is the constant term in the plane equation.

Plugging in the values, we get:

d = |(1(3) + 1(0) + 1(-2) - 2) / √(1² + 1² + 1²)|

d = |(1 + 0 - 4) / √(3)|

d = |-3 / √(3)|

d = |-√(3)|

Therefore, the shortest distance from the point (3, 0, −2) to the plane x + y + z = 2 is √(3) or approximately 1.732 units.

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The x-coordinate of the vertex of the parabola whose equation is given would be -3/2. Option C.

To find the x-coordinate of the vertex of the parabola, we need to use the formula:

In this case, a = 3 and b = 9, so:

x = -9/(2*3) = -3/2

Therefore, the x-coordinate of the vertex of the parabola is -3/2.

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The x-coordinate of the vertex of the parabola whose equation is given would be -3/2. Option C.

To find the x-coordinate of the vertex of the parabola, we need to use the formula:

In this case, a = 3 and b = 9, so:

x = -9/(2*3) = -3/2

Therefore, the x-coordinate of the vertex of the parabola is -3/2.

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Answer:

26.95

Step-by-step explanation:

pass =  656.26 = (36 s + 2t) so 17.27 per person assuming teacher & student same price.

lunch = 348.48/36 =9.68/student

pass and lunch = 9.68 + 17.27 =26.95

The bicycle travels 12 feet in 3 seconds.

This can be found by integrating the velocity function v(t) over the interval [0,3]: ∫4t dt = 2t² evaluated at t=3.

The velocity function v(t) gives the rate of change of distance with respect to time, so to find the total distance traveled over a given time interval, we need to integrate v(t) over that interval.

In this case, we want to find the distance traveled in 3 seconds, so we integrate v(t) from t=0 to t=3: ∫4t dt = 2t² evaluated at t=3 gives us the total distance traveled, which is 12 feet. This means that after 3 seconds, the bike has traveled 12 feet from its starting point.

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The Laplace transform of a+bt+c is (a/s) + (b/s^2) + (c/s). The Laplace transform of A cos(t+Bsin(t)) is (s/(s^2+B^2)) (A cos(φ) + (B/sin(φ)) A sin(φ)), where φ = arctan(B/s).

For a function f(t), the Laplace transform F(s) is defined as ∫[0, ∞) e^(-st) f(t) dt, where s is a complex number.

To find the Laplace transform of a+bt+c, we use linearity and the Laplace transform of elementary functions:

L{a+bt+c} = L{a} + L{bt} + L{c} = a/s + bL{t} + c/s = a/s + b/s^2 + c/s

Therefore, the Laplace transform of a+bt+c is (a/s) + (b/s^2) + (c/s).

B. To find the Laplace transform of A cos(t+Bsin(t)), we use the following identity:

cos(t + Bsin(t)) = cos(t)cos(Bsin(t)) - sin(t)sin(Bsin(t))

Then, we apply the Laplace transform to both sides and use linearity and the Laplace transform of elementary functions:

L{cos(t + Bsin(t))} = L{cos(t)cos(Bsin(t))} - L{sin(t)sin(Bsin(t))}

Using the formula L{cos(at)} = s/(s^2 + a^2), we get:

L{cos(t + Bsin(t))} = (s/(s^2+B^2)) L{cos(t)} - (s/(s^2+B^2)) L{sin(t)}

Using the formula L{sin(at)} = a/(s^2 + a^2), we get:

L{cos(t + Bsin(t))} = (s/(s^2+B^2)) (1/s) - (B/(s^2+B^2)) (1/s)

Simplifying, we get:

L{cos(t + Bsin(t))} = (s/(s^2+B^2)) (A cos(φ) + (B/sin(φ)) A sin(φ)), where φ = arctan(B/s)

Therefore, the Laplace transform of A cos(t+Bsin(t)) is (s/(s^2+B^2)) (A cos(φ) + (B/sin(φ)) A sin(φ)), where φ = arctan(B/s).

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The minimum pumping length of {} is any positive integer, of {a} is 1, {a, aaaa, aa}: 1, σ∗: 1 and of {ϵ} is not regular

To find the minimum pumping length for a given language, we need to consider the smallest possible strings in the language and find the smallest length at which we can apply the pumping lemma.

{} (the empty language): There are no strings in the language, so the pumping lemma vacuously holds for any pumping length. The minimum pumping length is any positive integer.

{a}: The smallest string in the language is "a". We can choose the pumping length to be 1, since any substring of "a" of length 1 is still "a". Thus, the minimum pumping length is 1.

{a, aaaa, aa}: The smallest string in the language is "a". We can choose the pumping length to be 1, since any substring of "a" of length 1 is still "a". Thus, the minimum pumping length is 1.

σ∗ (the Kleene closure of σ): Any string over {a} is in the language, including the empty string. We can choose the pumping length to be 1, since any substring of any string in the language of length 1 is still in the language. Thus, the minimum pumping length is 1.

{ϵ} (the language containing only the empty string): The smallest string in the language is the empty string, which has length 0. However, the pumping lemma requires that the pumping length be greater than 0. Since there are no other strings in the language, we cannot satisfy the pumping lemma for any pumping length. Thus, the language {ϵ} is not regular.

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In this sample, 154 college graduates and 132 non-college graduates support drilling for oil and natural gas off the coast of California. Therefore, the percentage of college graduates who support drilling is (154/438) x 100 = 35.16%, while the percentage of non-college graduates who support drilling is (132/389) x 100 = 33.95%.

It is worth noting that college graduates have a larger proportion of support than non-college graduates, although the difference is not statistically significant. The percentages of those who oppose and those who are unsure, on the other hand, differ dramatically between the two categories. In this sample, 41.1% of college graduates were opposed to drilling, compared to 32.4% of non-college graduates, and 23.7% were uncertain, compared to 33.6% of non-college graduates.

Overall, the evidence reveals that, while there is some difference in beliefs between college graduates and non-college graduates, the differences are not statistically significant. In both categories, the percentages of support, opposition, and undecided are quite identical. It is worth noting, however, that a sizable proportion of both groups (about one-third) are undecided, implying that there is still substantial disagreement and confusion around this subject.

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Answer:

35

Step-by-step explanation:

The overall reliability of the system through time t is approximately 0.370.

You have a system with four components and their reliabilities through time t are given as follows:

1. Component 1: 0.80
2. Component 2: 0.66
3. Component 3: 0.78
4. Component 4: 0.89

To find the overall reliability of the system, you'll need to multiply the reliabilities of each individual component:

Overall Reliability = Component 1 Reliability × Component 2 Reliability × Component 3 Reliability × Component 4 Reliability

Step-by-step calculation:

Overall Reliability = 0.80 × 0.66 × 0.78 × 0.89

Now, multiply the given reliabilities:

Overall Reliability ≈ 0.370

So, the overall reliability of the system through time t is approximately 0.370.

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a) U(X) is a nonzero function that satisfies E[U(X)] = 0, which shows that the family fo(x) = 2 is not complete.

b) U(X) is a nonzero function that satisfies E[U(X)] = 0, which shows that the family N(0,0) is not complete.

A nonzero function is a mathematical function that takes at least one value different from zero within its domain. In other words, there exists at least one input value for which the output value is not equal to zero.

i) To show that the family fo(x) = 2 is not complete, we need to find a nonzero function U(X) such that E[U(X)] = 0 for all e > 0. Let U(X) be defined as:

U(X) = { -1 if -4 < X < 0

1 if 0 < X < 4

0 otherwise

Then, we have:

E[U(X)] = ∫fo(x)U(x)dx = 2 ∫U(x)dx = 2 [∫(-4,0)-1dx + ∫(0,4)1dx] = 2(-4+4) = 0

Thus, U(X) is a nonzero function that satisfies E[U(X)] = 0, which shows that the family fo(x) = 2 is not complete.

ii) To show that the family N(0,0) is not complete, we need to find a nonzero function U(X) such that E[U(X)] = 0 for all e > 0. Let U(X) be defined as:

U(X) = X

Then, we have:

E[U(X)] = E[X] = ∫N(0,0)xdx = 0

Thus, U(X) is a nonzero function that satisfies E[U(X)] = 0, which shows that the family N(0,0) is not complete.

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The probability of x ≤ 8 successes in 10 trials of a binomial experiment with probability of success p = 0.6 is option (c) 0.954.

We can use the cumulative distribution function (CDF) of the binomial distribution to calculate the probability of getting x ≤ 8 successes in 10 trials with a probability of success p = 0.6.

The CDF gives the probability of getting at most x successes in n trials, and is given by the formula

F(x) = Σi=0 to x (n choose i) p^i (1-p)^(n-i)

where (n choose i) represents the binomial coefficient, and is given by

(n choose i) = n! / (i! (n-i)!)

Plugging in the values, we get

F(8) = Σi=0 to 8 (10 choose i) 0.6^i (1-0.6)^(10-i)

Using a calculator or a software program, we can calculate this as

F(8) = 0.9544

So the probability of getting x ≤ 8 successes is 0.9544.

Therefore, the answer is (c) 0.954.

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The type of r.v.R is a continuous random variable, since its possible values form a continuous interval [0,1].

The sample space of R is the interval [0,1], since the distance from the origin to any point inside the unit circle is between 0 and 1.
To find P(R < r), we need to find the probability that the randomly selected point falls inside a circle of radius r centered at the origin. The area of this circle is πr^2, and the area of the entire unit circle is π, so the probability is P(R < r) = πr^2/π = r^2.
The cdf of R is the function F(r) = P(R ≤ r) = ∫0r 2πx dx / π = r^2, where the integral is taken over the interval [0,r]. This is because the probability that R is less than or equal to r is the same as the probability that the randomly selected point falls inside the circle of radius r centered at the origin, which has area πr^2. The cdf of R is a continuous and increasing function on the interval [0,1].

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From the Venn diagram, the values of a, b, c and d are11,19,35,35 respectively

A Venn diagram is an illustration that uses circles to show the relationships among things or finite groups of things. Circles that overlap have a commonality while Circles that do not overlap do not share those traits.

The universal set is ∈ = 100

The languages are

Spanish = 30

Russian = 54

(S∪ R)¹ = 35 = d

a = Spanish only = a-b

30-b = a

Russia only = c-b

54 - b

Therefore, The universal set ∈ is

100 = (a-b) + (b)+ (c-b) +(d)

100 =  30-b + b + 54 - b + 35

100 = 119 - b = 119-100

b= 19

Therefore,

a = 30 -19 =11

b = 19

c = 59 - 19 35

d = 35

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1. Identify the linear equation. y = mx + b
2. Take (b) and plot it on the y axis. Since b is a positive 3, that means you plot a positive 3 on the y axis. This will be the number that your line crosses the y axis on.
3. Take (mx) and plot it in correlation to (b). mx = -x also known as -1. So, from +3 on the y axis, move once to the left and once down. Your coordinate should land on (2, 1).

From here on out, keep moving -1 on the y axis and +1 on the x axis. The ongoing coordinates should look something like (1, 2)(0, 3)(-1, 4) and so on.