A project has the following projected outcomes in dollars: $240, $310, and $560. The probabilities of their outcomes are 20%, 60%, and 20% respectively. What is the expected value of these outcomes?

Answer:

Equation is a quadratic equation (polynomial ax+by+c with highest degree 2), so it has two solutions.

Answer:

B. is correct because it only can go 1 way.

Step-by-step explanation:

Answer:

x= 114, y= 66

Step-by-step explanation:

Opposite angles of a parallelogram are equal

Answer:

3x² – 5x + 2 is a polynomial because:

Exponents are whole numbers, and the expression has at least 1 term.

Exponents other than whole numbers can take the form of variables in denominators, and roots which we don't want.

The scalar projection of b onto a is 5/9, and the vector projection of b onto a is (20/81, 35/81, -20/81).

To find the scalar and vector projections of vector b onto vector a, we can use the following formulas:

Scalar Projection:

The scalar projection of b onto a is given by the formula:

Scalar Projection = |b| * cos(θ)

Vector Projection:

The vector projection of b onto a is given by the formula:

Vector Projection = Scalar Projection * (a / |a|)

where |b| represents the magnitude of vector b, θ is the angle between vectors a and b, a is the vector being projected onto, and |a| represents the magnitude of vector a.

Let's calculate the scalar and vector projections of b onto a:

a = (4, 7, -4)

b = (4, -1, 1)

First, we calculate the magnitudes of vectors a and b:

|a| = √(4² + 7² + (-4)²) = √(16 + 49 + 16) = √81 = 9

|b| = √(4² + (-1)² + 1²) = √(16 + 1 + 1) = √18

Next, we calculate the dot product of vectors a and b:

a · b = (4 * 4) + (7 * -1) + (-4 * 1) = 16 - 7 - 4 = 5

Using the dot product, we can find the angle θ between vectors a and b:

cos(θ) = (a · b) / (|a| * |b|)

cos(θ) = 5 / (9 * √18)

Now, we can calculate the scalar projection:

Scalar Projection = |b| * cos(θ)

Scalar Projection = √18 * (5 / (9 * √18)) = 5 / 9

Finally, we calculate the vector projection:

Vector Projection = Scalar Projection * (a / |a|)

Vector Projection = (5 / 9) * (4, 7, -4) / 9 = (20/81, 35/81, -20/81)

Therefore, the scalar projection of b onto a is 5/9, and the vector projection of b onto a is (20/81, 35/81, -20/81).

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Answer:

let's divide the figure into two parts.

radius of the semicircle is 3.5m. two semi-circles make a circle and

area of circle=pi×r²

area of circle=22/7×(3.5m)2².

area of circle=38.5m²

area of rectangle=length ×width

area of rectangle =18m×7m

area of rectangle =126m²

area pf figure =38.5m²+126m²

area of figure=164.5m²

Answer:

reflection??.........

Answer:

they are congruent

Step-by-step explanation:

because they are the same size and have the smae area!

9514 1404 393

Answer:

  6 3/10 pounds

Step-by-step explanation:

The weight change will be found by multiplying the rate of change by the time.

  ∆w = (-1.8 lb/h)(3.5 h) = -6.3 lb

The total change in weight after 3 1/2 hours is 6 3/10 = 6.3 pounds.

The orange divisor(s) above are the prime factors of the number 7,776. If we put all of it together we have the factors 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 = 7,776. It can also be written in exponential form as 25 x 35.

Answer:

Step-by-step explanation:

As each step has the same depth and rise, they are respectively 1.2/4=0.3m and 1.8/4=0.45m.

Dividing the steps along the dotted lines, the total rise of the 4 concrete steps  = (1+2+3+4)*0.45

= 4.5m

Total concrete volume = total rise * depth * width

= 4.5*0.3*1.8

= 2.43m^3

Answer:30

Step-by-step explanation:

[tex]\sinh'(x) = \cosh(x)$.[/tex]

Hence, we have shown that [tex]\cosh'(x) = \sinh(x)$ and $\sinh'(x) = \cosh(x)$[/tex] using the rules for differentiating [tex]e^x$.[/tex]

What are Hyperbolic Functions?

Hyperbolic functions are a set of mathematical functions that are analogs of the trigonometric functions. While trigonometric functions are defined based on the unit circle, hyperbolic functions are defined using the hyperbola.

To show that [tex]\cosh'(x) = \sinh(x)$ and $\sinh'(x) = \cosh(x)$[/tex] using the rules for differentiating [tex]e^x$:[/tex]

[tex]\textbf{1. Derivative of $\cosh(x)$:}[/tex]

Starting with the definition of [tex]\cosh(x)$:[/tex]

[tex]\[\cosh(x) = \frac{1}{2}(e^x + e^{-x})\][/tex]

Taking the derivative with respect to x using the chain rule and the derivative of [tex]e^x$:[/tex]

[tex]\cosh'(x) &= \frac{1}{2}\left(\frac{d}{dx}(e^x) + \frac{d}{dx}(e^{-x})\right) \\\\&= \frac{1}{2}(e^x - e^{-x}) \\\\&= \sinh(x)[/tex]

Therefore, [tex]\cosh'(x) = \sinh(x)$.[/tex]

[tex]\textbf{2. Derivative of $\sinh(x)$:}[/tex]

Starting with the definition of [tex]\sinh(x)$:[/tex]

[tex]\[\sinh(x) = \frac{1}{2}(e^x - e^{-x})\][/tex]

Taking the derivative with respect to x using the chain rule and the derivative of [tex]$e^x$[/tex]:

[tex]\sinh'(x) &= \frac{1}{2}\left(\frac{d}{dx}(e^x) - \frac{d}{dx}(e^{-x})\right) \\\\&= \frac{1}{2}(e^x + e^{-x}) \\\\&= \cosh(x)[/tex]

Therefore, [tex]\sinh'(x) = \cosh(x)$.[/tex]

Hence, we have shown that [tex]\cosh'(x) = \sinh(x)$ and $\sinh'(x) = \cosh(x)$[/tex] using the rules for differentiating [tex]e^x$.[/tex]

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Answer:

Inequality Form:

h > 6

Interval Notation:

( 6 , ∞ )

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

The sum numbers are 1, 2, 4, 5, 8, 10, 20, 40

The approximation of the integral ∫cos(x³ - 5) dx using composite Simpson's rule with n = 3 is approximately 1.01259.

The integral ∫cos(x³ - 5) dx using composite Simpson's rule with n = 3, we need to divide the integration interval into smaller subintervals and apply Simpson's rule to each subinterval.

The formula for composite Simpson's rule is

I ≈ (h/3) × [f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + ... + 2f([tex]x_{n-2}[/tex]) + 4f([tex]x_{n-1}[/tex]) + f([tex]x_{n}[/tex])]

where h is the step size, n is the number of subintervals, and f([tex]x_{i}[/tex]) represents the function value at each subinterval.

In this case, n = 3, so we will have 4 equally-sized subintervals.

Let's assume the lower limit of integration is a and the upper limit is b. We can calculate the step size h as (b - a)/n.

Since the limits of integration are not provided, let's assume a = 0 and b = 1 for simplicity.

Using the formula for composite Simpson's rule, the approximation becomes:

I ≈ (h/3) [f(x₀) + 4f(x₁) + 2f(x₂) + 4f(x₃) + f(x₄)]

For n = 3, we have four equally spaced subintervals:

x₀ = 0, x₁ = h, x₂ = 2h, x₃ = 3h, x₄ = 4h

Using these values, the approximation becomes:

I ≈ (h/3) × [f(0) + 4f(h) + 2f(2h) + 4f(3h) + f(4h)]

Substituting the function f(x) = cos(x³ - 5):

I ≈ (h/3)  [cos((0)³ - 5) + 4cos((h)³ - 5) + 2cos((2h)³ - 5) + 4cos((3h)³ - 5) + cos((4h)³ - 5)]

Now, we need to calculate the step size h and substitute it into the above expression to find the approximation. Since we assumed a = 0 and b = 1, the interval width is 1.

h = (b - a)/n = (1 - 0)/3 = 1/3

Substituting h = 1/3 into the expression:

I ≈ (1/3)  [cos((-1)³ - 5) + 4cos((1/3)³ - 5) + 2cos((2/3)³ - 5) + 4cos((1)³ - 5) + cos((4/3)³ - 5)]

Evaluating the expression further:

I ≈ (1/3)  [cos(-6) + 4cos(-4.96296) + 2cos(-4.11111) + 4cos(-4) + cos(-3.7037)]

Approximating the values using a calculator, we get:

I ≈ 1.01259

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z - 20x - 12y + 76 = 0 is the required equation of the plane that is tangent to the given surface at the point (5, 3, – 60).

Given the function is z=8-2x²-2y² and point is (5, 3, – 60).

We need to find the equation of the plane tangent to the given surface at the given point. The gradient vector of the function f(x, y, z) is given by(∂f/∂x)i + (∂f/∂y)j + (∂f/∂z)k∂f/∂x= -4x and ∂f/∂y= -4y

Therefore, the gradient vector is given by-4xi -4yj + k

Therefore, the equation of the tangent plane is given byz - z1=∇f(x1, y1) . (x - x1)i + ∇f(x1, y1) . (y - y1)j + (-1) [f(x1, y1, z1)]

where (x1, y1, z1) is the given pointWe have f(5, 3, – 60) = 8 – 2(5²) – 2(3²) = – 60

Therefore, the equation of the plane is given byz + 60= (-20i - 12j + k) . (x - 5) - (16i + 24j + k) . (y - 3)

Thus, z - 20x - 12y + 76 = 0 is the required equation of the plane that is tangent to the given surface at the point (5, 3, – 60).

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Answer:

1.44m^3

Step-by-step explanation:

Given data

Number of balls= 8

Diameter of ball = 70cm = 0.7m

Radius= 35cm= 0.35m

We know that a ball has a spherical shape

The volume of a sphere is

V= 4/3πr^3

substitute

V= 4/3*3.142*0.35^3

V= 0.18m^3

Hence if 1 ball has a volume of 0.18m^3

Then 8 balls will have a volume of

=0.18*8

=1.44m^3

Answer:

42.09 cubic units

Step-by-step explanation:

[tex]\frac{4.11*5.12}{2} *4[/tex]

=42.0864, which rounds to 42.09

Note: The 6.57 is not needed to solve this problem

The vector that is parallel to the line of intersection of the planes 4x - 3y + 2z = 12 and x + 5y - z = 7 is option (c) -7i + 6j + 23k.

To find a vector that is parallel to the line of intersection of two planes, we need to determine the direction of the line. This can be achieved by finding the cross product of the normal vectors of the planes.

The normal vector of the first plane 4x - 3y + 2z = 12 is (4, -3, 2), obtained by taking the coefficients of x, y, and z. Similarly, the normal vector of the second plane x + 5y - z = 7 is (1, 5, -1).

To find the cross product of these two normal vectors, we take their determinant: (4i, -3j, 2k) x (1i, 5j, -1k). Evaluating the determinant, we get (-23i - 6j - 13k).

The resulting vector -23i - 6j - 13k is parallel to the line of intersection of the planes. However, the given options only include positive coefficients for i, j, and k. To match the given options, we can multiply the vector by -1 to obtain a parallel vector. Thus, -(-23i - 6j - 13k) simplifies to -7i + 6j + 23k, which matches option (c). Therefore, option (c) -7i + 6j + 23k is the vector parallel to the line of intersection of the planes.

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We can be 90% confident that the true difference between the mean daily calorie consumption of females in the September-February period and the mean daily calorie consumption of females in the March-August period falls within the range of 21460.3 to 23033.7 calories.

In this study conducted by a fitness magazine, two separate samples of females were chosen to investigate the difference in mean daily calorie consumption between two time periods: September-February and March-August. The first sample consisted of 265 females, and the second sample consisted of 220 females. The mean daily calorie consumption and standard deviations were calculated for each period. This information will be used to construct a confidence interval to estimate the difference between the mean daily calorie consumption of females in the two periods.

To construct a confidence interval for the difference between the mean daily calorie consumption of females in the September-February and March-August periods, we can use the formula:

Confidence Interval = (X₁ - X₂) ± (Z * SE)

Where:

X₁ and X₂ are the sample means of the two periods (September-February and March-August, respectively)

Z is the critical value associated with the desired confidence level (90% confidence level corresponds to Z = 1.645)

SE is the standard error of the difference between the means

First, let's calculate the sample means and standard deviations for each period:

For the September-February period: X₁ = 23873 calories, σ₁ = 192 (standard deviation), n₁ = 265 (sample size)

For the March-August period: X₂ = 2412.7 calories, σ₂ = 237.5 (standard deviation), n₂ = 220 (sample size)

Next, we calculate the standard error (SE) of the difference between the means using the formula:

SE = √((σ₁² / n₁) + (σ₂² / n₂))

Substituting the given values, we have:

SE = √((192² / 265) + (237.5² / 220))

Now, we can calculate the confidence interval using the formula mentioned earlier. With a 90% confidence level, the critical value Z is 1.645.

Substituting in the values, we get:

Confidence Interval = (23873 - 2412.7) ± (1.645 * SE)

Substituting the calculated value of SE, we can find the confidence interval:

Confidence Interval = (21460.3, 23033.7)

Therefore, we can be 90% confident that the true difference between the mean daily calorie consumption of females in the September-February period and the mean daily calorie consumption of females in the March-August period falls within the range of 21460.3 to 23033.7 calories.

Note: The confidence interval represents a range of values within which we believe the true difference lies, based on the given data and the selected confidence level.

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Answer:

0.43496 miles

Step-by-step explanation:

To convert from km to miles you can divide the km by 1.609 and that should give you an aproximate value for miles.

Answer: D

Step-by-step explanation:

The solution to the given initial-value problem is a power series given by y(x) = 1 - 2x^3 + 3x^4 + O(x^5).  As x increases, higher powers of x become significant, and the series must be truncated at an appropriate order to maintain accuracy .

y(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + ..., where a_0, a_1, a_2, ... are constants to be determined. We then differentiate the series term-by-term to find the derivatives y' and y''. Differentiating y(x), we have

y' = a_1 + 2a_2x + 3a_3x^2 + 4a_4x^3 + ..., and differentiating once more, we find y'' = 2a_2 + 6a_3x + 12a_4x^2 + ...Substituting these expressions into the given differential equation, we have:

(x + 3)(2a_2 + 6a_3x + 12a_4x^2 + ...) + 2(a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + ...) = 0

Given the initial conditions y(0) = 1 and y'(0) = 0, we can use these conditions to find the values of a_0 and a_1. Plugging in x = 0 into the power series, we have a_0 = 1. Differentiating y(x) and evaluating at x = 0, we get a_1 = 0.Therefore, the power series solution is y(x) = 1 + a_2x^2 + a_3x^3 + a_4x^4 + ..., where a_2, a_3, a_4, ... are yet to be determined.

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The average sum of differences of a series of numerical data from their mean is zero (option a).

This property holds true for any data set when calculating the mean deviation (also known as the average deviation) from the mean. The mean deviation is calculated by taking the absolute difference between each data point and the mean, summing them up, and dividing by the number of data points.

However, it's important to note that this property does not hold true when using squared differences, which is used in the calculation of variance and standard deviation. In those cases, the average sum of squared differences from the mean would give the variance (option c) or the squared standard deviation (option e).

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