Using Rolle’s Theorem find the two x-intercepts of the function f and show that f(x) = 0 at some point between the two x-intercepts. f(x) = x^2 - x - 2

Answer:99% of 110 is 108.9

so your answer is 108.9

Step-by-step explanation:

Because I said so

The equation of the tangent plane to the surface z=ly2-922 at the point (1,-3,0) is z = -27 - 6ly(x-1).

To find the equation of the tangent plane to the surface z=ly2-922 at the point (1,-3,0), we need to first find the partial derivatives of z with respect to x and y.

∂z/∂x = 0

∂z/∂y = 2ly

Next, we need to evaluate these partial derivatives at the given point (1,-3,0), which gives us:

∂z/∂x(1,-3,0) = 0

∂z/∂y(1,-3,0) = -6l

Using these partial derivatives, we can now write the equation of the tangent plane:

z = f(a,b) + ∂f/∂x(a,b)(x-a) + ∂f/∂y(a,b)(y-b)

where f(a,b) is the value of the function at the point (a,b), and (x-a) and (y-b) are the differences between the given point and the point on the tangent plane.

Plugging in the values we found, we get:

z = -27 - 6ly(x-1)

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The probability of B given A is approximately 24.9%.

The probability is typically defined as the proportion of positive outcomes to all outcomes in the sample space. Probability of an event P(E) = (Number of positive outcomes) (Sample space) is how it is written.

We can use the formula for conditional probability to solve for P(B | A):

P(B | A) = P(A & B) / P(A)

Substituting the given values, we get:

P(B | A) = 4.2% / 16.9%

        = 0.2485207100591716

To convert this to a percentage rounded to the nearest tenth, we can multiply by 100 and round to one decimal place:

P(B | A) ≈ 24.9%

Therefore, the probability of B given A is approximately 24.9%.

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The value of RE for the given triangle is 3 units.

A key idea in geometry known as the Pythagorean theorem outlines the relationship between the sides of a right triangle. It asserts that the square of the length of the longest side, known as the hypotenuse, is equal to the sum of the squares of the two shorter sides of a right triangle. In other words, if a and b are the measurements of a right triangle's two shorter sides and c is the measurement of the hypotenuse,

Triangle DRF is an right angle triangle.

Using Pythagoras Theorem we have:

c² = a² + b²

DF² = 27² + 9²

DF² = 729 + 81

DF² = 810

DF = √(810)

DF = 9 √(10)

Now, the value of DF = 9 sqrt(10).

In triangle DEF we have:

DE² = (9 √(10))² + (3 √(10))

DE² = 81 * 10 + 9 * 10

DE² = 900

DE = 30

Now, DE = DR + RE

Given, DR = 27 substituting the value:

Thus,

30 = 27 + RE

RE = 3

Hence, the value of RE for the given triangle is 3 units.

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To convert the polar equation T= 3sin θ + 5cos θ into a Cartesian equation, we can use the following identities:

sin θ = y / r
cos θ = x / r

where r is the distance from the origin to the point (x, y) in the Cartesian plane. Substituting these expressions into the given polar equation, we get:

T = 3(y / r) + 5(x / r)

Multiplying both sides by r, we obtain:

Tr = 3y + 5x

This is the Cartesian equation of the given polar equation.

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There are 46,620 number of ways to choose three different members to be president, vice president, and treasurer from a club with 20 women and 17 men.

1: Calculate the total number of members in the club.

Total members = Women + Men = 20 + 17 = 37

2: Determine the number of ways to choose the president.

Since there are 37 members, there are 37 options for president.

3: Determine the number of ways to choose the vice president.

After the president has been chosen, there are 36 remaining members to choose from for vice president.

4: Determine the number of ways to choose the treasurer.

After choosing both the president and vice president, there are 35 remaining members to choose from for treasurer.

Step 5: Calculate the total number of ways to choose the three different members.

Total ways = Ways to choose president × Ways to choose vice president × Ways to choose treasurer = 37 × 36 × 35 = 46,620

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The probability that the z-score is greater than 2.064. Using a standard normal table or calculator, we find that the probability is approximately 1.97%. This means there is a 1.97% chance that the average weight of the fish caught by James would be greater than 3 pounds.

To solve this problem, we can use the central limit theorem, which states that the distribution of the sample means approaches a normal distribution as the sample size increases, regardless of the shape of the population distribution. We can also use the z-score formula to standardize the sample mean.

First, we need to calculate the standard error of the mean (SEM) using the formula:

SEM = standard deviation / square root of sample size
SEM = 1.211 / square root of 25
SEM = 0.2422

Next, we can calculate the z-score using the formula:

z = (sample mean - population mean) / SEM
z = (3 - 2.5) / 0.2422
z = 2.066

Finally, we can use a standard normal distribution table (or a calculator that has a normal distribution function) to find the probability of getting a z-score greater than 2.066. The probability is approximately 0.0197 or 1.97%.

Therefore, the probability that the average weight of the fish caught by James would be greater than 3 pounds is 1.97%.
To answer your question, we need to use the normal distribution and the given information to calculate the probability. Here are the terms you mentioned:

- Smallmouth bass: a type of freshwater fish
- Mean: 2.5 pounds
- Standard deviation: 1.211 pounds
- Number of fish caught by James: 25
- We need to find the probability that the average weight is greater than 3 pounds.

To solve this, we can use the z-score formula for the sample mean:

z = (X - μ) / (σ / √n)

Where:
- X is the sample mean (3 pounds)
- μ is the population mean (2.5 pounds)
- σ is the population standard deviation (1.211 pounds)
- n is the sample size (25 fish)

Calculating the z-score:

z = (3 - 2.5) / (1.211 / √25)
z = 0.5 / (1.211 / 5)
z = 0.5 / 0.2422
z ≈ 2.064

Now, we need to find the probability that the z-score is greater than 2.064. Using a standard normal table or calculator, we find that the probability is approximately 1.97%. This means there is a 1.97% chance that the average weight of the fish caught by James would be greater than 3 pounds.

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a. The accumulation function a(t) applies to real numbers t > 0

b. The accumulation function a(t) applies to real numbers t > 0

c. For any 0 < t < 1, the accumulation function a(t) approaches 0 as t approaches 0.

d. For any t > 1, the accumulation function a(t) is proportional to 2t and approaches 2t * f(c) as t approaches infinity.

a. To prove that the accumulation function a(t) applies to real numbers t > 0, we need to show that it makes sense to evaluate the function for any positive real value of t.
The accumulation function a(t) is defined as the integral from 0 to t of some function f(x). Since integrals are defined for any continuous function over a closed interval, we know that f(x) must be a continuous function over the interval [0, t].
Furthermore, the domain of integration is the closed interval [0, t], which includes all real numbers between 0 and t (including 0 and t themselves). Therefore, the accumulation function a(t) applies to all real numbers t > 0.
b. This is the same as part a, so the answer is the same: the accumulation function a(t) applies to all real numbers t > 0.
c. To prove that for 0 < t < 1, then ..., where t is a real number, we need to evaluate the accumulation function a(t) for values of t between 0 and 1.
Let's assume that f(x) is a continuous function over the interval [0, 1]. Then, the accumulation function a(t) is given by:
a(t) = ∫[0,t] f(x) dx
Since 0 < t < 1, we know that the interval [0, t] is a subset of the interval [0, 1]. Therefore, we can use the Mean Value Theorem for Integrals to say that there exists some value c in [0, t] such that:
a(t) = t * f(c)
Since f(x) is a continuous function over [0, 1], we know that it attains a maximum value of M and a minimum value of m over the interval [0, 1]. Therefore, we can say that:
m * t ≤ a(t) ≤ M * t
This means that the accumulation function a(t) is bounded between two values that are proportional to t. As t approaches 0, these bounds approach 0 as well, so we can say that:
lim(t → 0) a(t)/t = 0
Therefore, for any 0 < t < 1, the accumulation function a(t) approaches 0 as t approaches 0.
d. To prove that for t > 1, then ..., where t is a real number, we need to evaluate the accumulation function a(t) for values of t greater than 1.
Let's assume that f(x) is a continuous function over the interval [0, t]. Then, the accumulation function a(t) is given by:
a(t) = ∫[0,t] f(x) dx
Since t > 1, we know that the interval [0, t] is a subset of the interval [0, 2t]. Therefore, we can use the Mean Value Theorem for Integrals to say that there exists some value c in [0, 2t] such that:
a(t) = ∫[0,t] f(x) dx = ∫[0,2t] f(x) dx - ∫[t,2t] f(x) dx = 2t * f(c) - ∫[t,2t] f(x) dx
Since f(x) is a continuous function over the interval [0, t], we know that it attains a maximum value of M and a minimum value of m over this interval. Therefore, we can say that:
m * t ≤ ∫[t,2t] f(x) dx ≤ M * t
This means that the second term in the equation for a(t) is bounded between two values that are proportional to t. As t approaches infinity, these bounds approach 0 as well, so we can say that:
lim(t → ∞) ∫[t,2t] f(x) dx/t = 0
Therefore, for any t > 1, the accumulation function a(t) is proportional to 2t and approaches 2t * f(c) as t approaches infinity.

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Two angles that form two pairs of opposing rays are called vertical angles. A triangle's total angles are always equal to 180 degrees.

Two rays that originate from the same location and go in completely different directions are said to be opposing rays. As a result, the two rays combine to form a single straight line that passes through the shared terminal Q2.

Two angles that form two pairs of opposing rays are called vertical angles. These might be considered as the opposing angles of an X.

A triangle's total angles are always equal to 180 degrees. This indicates that the outcome will be as follows if we sum the measurements of all three angles of a triangle

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The expected value of that Poisson distribution is 12.96. So, correct option is A.

The expected value of a Poisson distribution is given by lambda, where lambda is the mean or average number of occurrences in the given interval. In a Poisson distribution, the variance is equal to the mean, so the standard deviation is the square root of the mean.

Thus, if the standard deviation is known to be 3.60, we can find the mean as follows:

Standard deviation = √(mean)

Squaring both sides, we get:

Variance = mean

Substituting the given standard deviation, we have:

3.60 = √(mean)

Squaring both sides again, we get:

12.96 = mean

Therefore, correct option is A.

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The area of the region included between the parabolas:  [tex]A = 4 [(2/3)p^2\sqrt{p } + (2/3)\sqrt{p } - (2/5)p^2\sqrt{p}[/tex]

To find the area of the region between the parabolas [tex]y^2 = 4(p+1)(x + p+ 1)[/tex] and [tex]y^2 = 4(p^2 + 1)(p^2 + 1 - x),[/tex]  we need to first graph the two parabolas and determine the intersection points.

Then, we can integrate the difference between the y-coordinates of the two parabolas over the range of x where they intersect.

Let's start by graphing the two parabolas:

[tex]y^2 = 4(p+1)(x + p + 1)[/tex] is a parabola that opens to the right and has its vertex at the point (-p-1, 0).

The distance between the vertex and the focus is p+1, and the distance between the vertex and the directrix is also p+1.

[tex]y^2 = 4(p^2 + 1)(p^2 + 1 - x)[/tex] is a parabola that opens to the left and has its vertex at the point (p^2+1, 0).

The distance between the vertex and the focus is [tex]\sqrt{ (p^2+1) }[/tex], and the distance between the vertex and the directrix is also [tex]\sqrt{(p^2+1). }[/tex]

Now, we need to find the intersection points of the two parabolas. Setting the two equations equal to each other, we get:

[tex]4(p+1)(x+p+1) = 4(p^2+1)(p^2+1-x)[/tex]

Simplifying and rearranging, we get:

[tex]x = p^3 + p^2 - p - 1[/tex]

Substituting this value of x into either of the original equations, we get:

[tex]y^2 = 4(p+1)(p^2+2p)[/tex]

Simplifying, we get:

[tex]y^2 = 4p(p+1)^2[/tex]

Taking the square root of both sides, we get:

y = ± 2(p+1)√p

Therefore, the two parabolas intersect at the points [tex](p^3+p^2-p-1, 2(p+1)\sqrt{p } )[/tex] and [tex](p^3+p^2-p-1, -2(p+1)\sqrt{p} ).[/tex]

Now, we can find the area of the region between the parabolas by integrating the difference between the y-coordinates of the two parabolas over the range of x where they intersect.

This gives:

[tex]A = 2\int [p^3+p^2-p-1, p^2+1] [2(p+1)√p - 2\sqrt{(p(p+1)^2)} ] dx[/tex]

Simplifying, we get:

[tex]A = 4 \int [p^3+p^2-p-1, p^2+1] (p-1)\sqrt{p} dx[/tex]

Integrating with respect to x, we get:

[tex]A = 4 [(2/3)(p^2+1)\sqrt{p} - (2/5)(p^2+1)p^{3/2} + (1/2)(p^2+1)^{3/2} - (2/3)(p^3+p^2-p-1)\sqrt{p} + (2/5)(p^3+p^2-p-1)p^{3/2} - (1/2)(p^3+p^2-p-1)^{3/})][/tex]

Simplifying, we get:

[tex]A = 4 [(2/3)p^2\sqrt{p } + (2/3)\sqrt{p } - (2/5)p^2\sqrt{p}[/tex]

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The least value of k for which the alternating series error bound guarantees that [S-Pk}= 3/100 is k = 34. To get the least value of k for which the alternating series error bound guarantees that [S-Pk}= 3/100,

We need to use the formula for the alternating series error bound: |S-Pk| ≤ a(k+1)
where a is the absolute value of the term following the last one included in Pk. In this case, the last term included in Pk is the kth term, which is (-1)^(k+1) * 2/k. Therefore, a = 2/(k+1).
We want to find the least value of k such that a(k+1) ≤ 3/100. Substituting a and simplifying, we get: 2(k+1)/(k+2) ≤ 3/100
Multiplying both sides by (k+2) and simplifying, we get: 200k + 400 ≤ 3k^2 + 6k
Rearranging and simplifying, we get: 3k^2 - 194k - 400 ≥ 0
Solving this quadratic inequality using the quadratic formula, we get: k ≤ (194 + sqrt(4*3*400+194^2))/6 or k ≥ (194 - sqrt(4*3*400+194^2))/6
k ≤ 33.053 or k ≥ 63.947
Therefore, the least value of k for which the alternating series error bound guarantees that [S-Pk}= 3/100 is k = 34.

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The total mass is 125π/6.

The Center of Mass is (625/4, 0)

We have a disk

x² + y² ≤25

let (x', y') be the center of Mass

So, x' = [tex](\int\limits_D {x. \rho(x,y)} \, dA[/tex]  / M

and, y' =  [tex](\int\limits_D {y. \rho(x,y)} \, dA[/tex]  / M

where M is the mass of the region

Also, p(x,y) = 5(x2 + y²)

x= r cos [tex]\theta[/tex] and y= r sin [tex]\theta[/tex]

Then, (r cos [tex]\theta[/tex] )² + (r sin [tex]\theta[/tex])² ≤25

r² < 25

r≤5

So, the total Mass is

= ∫ 5(x² + y²) dA

= [tex]\int\limits^{\pi/2}_0 \int\limits^5_ 0{ 5 (r^2 cos^2 \theta + r^2 sin^2 \theta)} \, dr d\theta[/tex]

= [tex]\int\limits^{\pi/2}_0 \int\limits^5_ 0{ 5 r^2 } \, dr d\theta[/tex]

= 5 x 125/3 [tex]\int\limits^ {\pi/2} _0{ d\theta[/tex]

= 5³/3. π/2

= 125π/6

Now, x' = [tex](\int\limits_D {x. \rho(x,y)} \, dA[/tex]  / M

x' = [tex]\int\limits^{\pi/2}_0 \int\limits^5_ 0{ r cos\theta 5 r^2 } \, dr d\theta[/tex]

x' = 5.5³/4

x' = 625/4

and, y' = [tex]\int\limits^{\pi/2}_0 \int\limits^5_ 0{ r sin\theta 5 r^2 } \, dr d\theta[/tex]

y' = 0

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a) Note that the inequality  expressing the possible number of piccolos that could perform allowing the audience to hear every note  is given as: 3,850x ≤ 16,000

b) the maximum number of piccolos allowed to perform this piece is 4.16 or approximately 4.

A) Where the above conditions are given, let us call the number of piccolos "x" as the number that can perform in a way that the audience will hear all the notes pplayed.

Recall that specialized acoustic tiles that eliminate any sounds greater than 16 kHz that means, all the frequencies put together must  nevr exced 16kHz. Thus,

the inequality will be 3,850x ≤ 16,000


B) To determine the maximum number of piccolos:

we must solve for x in the above equation:

3,850x ≤ 16,000

we divide both sides by 3850

3,850x/3850 ≤ 16,000/3850
x ≤ 4.15584415584

Hence, since x cannot be be decimal, that is the number of picollows, we must approximate x to 4

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The values of x where f has a relative maximum are -1 and 3. The answer is D, -1 and 3 only.

To determine the values of x where the function f has a relative maximum, we need to analyze the sign of the derivative in the neighborhood of each critical point.

The critical points of f correspond to the values of x where the derivative is equal to zero or undefined. In this case, the derivative is undefined only at x = -1, where there is a vertical asymptote.

To find the values of x where the derivative is zero, we set f'(x) = 0 and solve for x:

[tex]f'(x) = x(x-3)^2[/tex](x+1) = 0

This equation is satisfied when x = 0 or x = 3, since these are the only values that make one of the factors equal to zero. Therefore, these are the only two critical points of f.

Next, we can use the first derivative test to determine the nature of each critical point. This involves checking the sign of the derivative in the interval to the left and right of each critical point.

For x = -1, the derivative is undefined to the left and right, but we can still examine the sign of f'(x) in the vicinity of -1. For example, if we consider values of x slightly to the left of -1 (say, x = -1.1), we see that f'(x) is negative. Similarly, if we consider values of x slightly to the right of -1 (say, x = -0.9), we see that f'(x) is positive. This means that f has a relative maximum at x = -1.

For x = 0, the derivative is negative to the left and positive to the right. This means that f has a relative minimum at x = 0.

For x = 3, the derivative is positive to the left and negative to the right. This means that f has a relative maximum at x = 3.

Therefore, the values of x where f has a relative maximum are -1 and 3. The answer is D, -1 and 3 only.

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a.In the Taylor series expansion the second-order partial derivatives of g at the point (1,2) are gxx = 2, gyy = -2, and gxy = 3.

b. The gradient of g at the point (1,2) is z = 2x - 2y + 1.

a. To find the second-order partial derivatives of g at (1,2), we take the partial derivatives of each term of the given polynomial expansion. The resulting second-order partial derivatives are d²g/dx²=2, d²g/dy²=-2, and d²g/dxdy=3.

b. To find the gradient of g at (1,2), we take the partial derivatives of g with respect to x and y and evaluate them at (1,2). The resulting gradient is ∇g(1,2) = <1, -1>.

To find the equation of the tangent plane to the surface defined by z=g(x,y) at (1,2,1), we use the point-normal form of a plane equation. The normal vector is the gradient of g at (1,2), so the equation of the tangent plane is (x-1) - (y-2) + z-1 = 0.

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The function g(x) is a reflection over the x-axis and stretch by 3 units and shifted upward by 1 unit of the parent function f(x).

We have,

A capability is a declaration, idea, or rule that lays out a relationship between two factors. Capabilities might be found all through science and are fundamental for the advancement of huge connections.

The functions are given below.

f(x) = x

g(x) = -3f(x) + 1

g(x) = -3x + 1

The parent function is a reflection over the x-axis and stretch by 3 units and shifted upward by 1 unit.

The functions g(x) and f(x) are shown below.

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To evaluate the integral ∫∫ D (4x+2) dA over the region D bounded by the curves y=x^2 and y=2x, we can use a double integral in terms of x and y:

∫∫ D (4x+2) dA = ∫[0,2]∫[y/2,√y] (4x+2) dx dy

where the limits of integration for x come from the intersection of the two curves y=x^2 and y=2x.

First, we integrate with respect to x:

∫[y/2,√y] (4x+2) dx = 2x^2 + 2x |[y/2,√y]

= 2y + y√y

Then, we integrate with respect to y:

∫[0,2] (2y + y√y) dy

= (2/3)y^3 + (2/5)y^(5/2) |[0,2]

= (2/3)(2)^3 + (2/5)(2)^(5/2)

= 16/3 + 8/5√2

Therefore, the value of the integral is 16/3 + 8/5√2.

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A vector space can have multiple bases. So the given statement is false.

A vector space can have more than one basis. A basis is a set of linearly independent vectors that spans the entire vector space. A vector space can have infinitely many bases, and each basis may contain a different number of vectors. However, all bases of a vector space will have the same cardinality, which is called the dimension of the vector space.

Therefore, a vector space can have multiple bases.

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a) Cost of n paperbacks = 8n

b) Cost of paperbacks for a Book Club member, including a membership is represented by algebraic equation = 10 + 6n

c) Difference = $14 (96 - 82), Book Club membership saves money.

Cost refers to the amount of money or resources that are required to produce, purchase, or obtain a product or service. It is the value given up in exchange for something.

An algebraic expression is a combination of variables, constants, and mathematical operations, such as addition, subtraction, multiplication, and division, that can be simplified and evaluated.

According to the given information:

a) The cost of n paperbacks can be represented by the algebraic expression: 8n. This expression means that the cost of n paperbacks is equal to the unit price of $8 multiplied by the number of paperbacks purchased.

b) The cost of paperbacks for a Book Club member, including a membership can be represented by the algebraic expression: 10 + 6n. This expression means that the cost of paperbacks for a Book Club member is equal to the membership fee of $10 plus the discounted unit price of $6 multiplied by the number of paperbacks purchased.

c) Without the Book Club membership, the cost of 12 books would be 12 x 8 = $96. With the Book Club membership, the cost of 12 books would be 10 (membership fee) + 6(12) = $82. Therefore, the difference between the cost of 12 books purchased with and without a membership is $14 (96 - 82). This shows that being a Book Club member results in a significant discount on the cost of paperbacks.

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For an exponential distribution of time (in years) until part failure for a certain car, probability that the time until the first critical-part failure is less than 1 year is equals to the 0.2684.

The exponential distribution is a type of continuous probability distribution that is used to measure the expected time for an event to occur. Formula for it is

[tex]f(X) = \lambda e^{-\lambda x} ; X>0[/tex], where [tex]\lambda [/tex]--> rate parameter

x --> observed value

We have time ( in year) first critical-part failure for a certain car is exponentially distributed. Let X be the time until the first critical part failure for a certain car. Now, X follows Exponential distribution with mean 3.2 years. The probability density function of X is [tex]f(X) = \lambda e^{ - \lambda x} ; X>0 [/tex]

Here in this problem, [tex] \lambda =\frac{1}{3.2} [/tex] = 0.3125

Using the formula the probability of X less than 1 is P(X< 1) = [tex]1 - e^{ - 0.3125×1}[/tex]

P(X< 1 ) = 1 - 0.745189

P(X<1)= 0.2684

The required probability is 0.2684.

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The probability that a randomly selected worker primarily drives a car or takes public transportation to work is 0.915. The probability that a randomly selected worker neither drives a car nor takes public transportation to work is 0.085. The probability that a randomly selected worker does not drive a car to work is 0.133.

(a) To find the probability that a randomly selected worker primarily drives a car or takes public transportation to work, we can add the probabilities of each event:
P(drives car) + P(takes public transportation) = 0.867 + 0.048 = 0.915
Therefore, the probability that a randomly selected worker primarily drives a car or takes public transportation to work is 0.915.
(b) The probability that a randomly selected worker neither drives a car nor takes public transportation to work can be found by subtracting the probability of driving a car or taking public transportation from 1:
P(neither drives car nor takes public transportation) = 1 - P(drives car or takes public transportation)
P(neither drives car nor takes public transportation) = 1 - 0.915
P(neither drives car nor takes public transportation) = 0.085
Therefore, the probability that a randomly selected worker neither drives a car nor takes public transportation to work is 0.085.
(e) To find the probability that a randomly selected worker does not drive a car to work, we can subtract the probability of driving a car from 1:
P(does not drive car) = 1 - P(drives car)
P(does not drive car) = 1 - 0.867
P(does not drive car) = 0.133
Therefore, the probability that a randomly selected worker does not drive a car to work is 0.133.

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The value of f(-1) is 2

A function is defined as a relation between a set of inputs having one output each. In simple words, a function is a relationship between inputs where each input is related to exactly one output. Every function has a domain and codomain or range. A function is generally denoted by f(x) where x is the input. The general representation of a function is y = f(x).

The function is :

[tex]f'(x) = 12x^2+12x[/tex]

We integrate the above function, we get :

[tex]f(x) = 4x^3+6x^2+c[/tex] ....(1)

That's f(x) . so when x = 2 and y = 10 inside this function.

10 = [tex]4(2)^3+6(2)^2+c[/tex]

10 = 32 + 24 + c

10 = 56 + c

10 - 56 = c

c = - 46

Again, We have to find the f(-1)

And, Plug the value of x = -1 in equation (1)

f(-1) = [tex]4(-1)^3+6(-1)^2+c[/tex]

f(-1) = -4 + 6

f(-1) = 2

The original function is therefore [tex]f(x) = 4x^3+6x^2+c[/tex].

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Rounding to three decimal places, we get 0.343 or approximately 0.222 when expressed as a percentage.

The probability that a Conservative voter opposed raising taxes, we need to look at the second row of the table, which shows the proportions of Conservative voters who favoured or opposed raising taxes.

The proportion who opposed raising taxes is 0.12.
Therefore, the answer is B. 0.222.
To calculate this probability, we simply divide the number of Conservative voters who opposed raising taxes by the total number of Conservative voters:
0.12 / (0.23 + 0.12) = 0.12 / 0.35 = 0.342857
Rounding to three decimal places, we get 0.343 or approximately 0.222 when expressed as a percentage.

This means that there is a 22.2% chance that a Conservative voter opposed raising taxes.

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The answer is D. East Hills HS is negatively skewed, and Southview HS is positively skewed.

In order to compare the shapes of the distributions, we need to look at the skewness of the distributions.

Skewness is a measure of the asymmetry of a probability distribution. A distribution is said to be negatively skewed if the tail on the left-hand side of the probability density function is longer or fatter than the right-hand side. Conversely, a distribution is said to be positively skewed if the tail on the right-hand side of the probability density function is longer or fatter than the left-hand side.

Based on the answer choices, we can eliminate options B and C, as they both indicate that one of the distributions is symmetric, which is not possible if the other is skewed.

Now, we need to determine which distribution is positively skewed and which is negatively skewed.

Option A indicates that East Hills HS is negatively skewed, and Southview HS is symmetric. This is not possible since a negatively skewed distribution cannot be symmetric.

Option D indicates that East Hills HS is negatively skewed, and Southview HS is positively skewed. This is a valid comparison since it is possible for one distribution to be negatively skewed while the other is positively skewed.

Therefore, the correct answer is D. East Hills HS is negatively skewed, and Southview HS is positively skewed.

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a. relative frequency table can be made by calculating the percentage of students who use each mode of transportation. We may get the answer by dividing the number of people using each technique by the total number of students

Methods          Number of People                 Relative Frequency

Carpool                         35                                              35/96

Drive                         14                                              14/96

Public transport         47                                              47/96

Total                         96                                                1

b. The likelihood that a student takes public transportation or a carpool to get to school is equal to the likelihood that they do not drive.

P(not driving) = P(carpooling) + P(public transport)

P(not driving) = 35/96 + 47/96

P(not driving) = 82/96

P(not driving) = 0.8542 or 85.42%

c. The likelihood that a student will either drive or participate in a carpool to get to school is equal to the likelihoods of both options.

P(carpooling or driving) = P(carpooling) + P(driving)

P(carpooling or driving) = 35/96 + 14/96

P(carpooling or driving) = 49/96

P(carpooling or driving) = 0.5104 or 51.04%

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C = -3, and the final expression for f(x) is:

f(x) = 2sinx + 9cosx - 3

To discover f(x) whilst f'(x) = 2cosx - 9sinx and f(0)=6, we need to integrate f'(x) with recognize to x to obtain f(x), whilst also considering the regular of integration.

∫f'(x) dx = ∫(2cosx - 9sinx) dx

the use of the integration rules for cosine and sine, we get:

f(x) = 2sinx + 9cosx + C

Wherein C is the constant of integration.

To discover the value of C, we use the given initial circumstance f(0) = 6:

f(0) = 2sin(0) + 9cos(0) + C = 9 + C = 6

Therefore, C = -3, and the final expression for f(x) is:

f(x) = 2sinx + 9cosx - 3

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Answer:

The answer is 100°

Step-by-step explanation:

supplementary angles are angles that sum up to 180°

x+80=180

x=180-80

x=100°

Answer:

100

Step-by-step explanation:

A supplementary angle have 180°

[tex]x + 80 = 180\\[/tex]

then

[tex]x = 100[/tex]

so the answer is

[tex]100[/tex]

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