I need some help on this one!!!

The 95% confidence interval as per given values is CI = (0.0315, 0.0605)

Total number of people who consume soft drinks weekly, = 11705

People who were smokers = 2527

Total number of people who dont consume soft drinks weekly = 14685

People who were smokers = 2503

Calculating the sample size -

p1 = 2527/11705

Smokers/ people consuming soft drinks weekly,

= 0.2159

p2 = 2503/14685

= 0.1703

Smokers/ people consuming soft drinks weekly,

[tex]SE = √[(p1(1-p1)/n1) + (p2(1-p2)/n2)][/tex]

Therefore,

[tex]SE = √[(0.2159 (1-0.2159)/11705) + (0.1703 (1-0.1703)/14685)][/tex]

= 0.0074

Calculating the confidence interval -

[tex]CI = (p1 - p2) ± z x SE[/tex]

Using z =1.96 for 95% confidence, the confidence interval is:

CI = (0.2159 - 0.1703) ± 1.96 x 0.0074

= 0.0456 ± 0.0145

= CI = (0.0315, 0.0605) ( After rounding to decimal places)

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The inverse Laplace transform of f(s) = -e^(-4s) (4s+6)/(s^2+6^2) is f(t) = -2e^(-2t) sin(6t) u(t).

To find the inverse Laplace transform of f(s), we first factor the denominator of f(s) to obtain f(s) = -e^(-4s) (4s+6)/[(s+3i)(s-3i)]. Then we use partial fraction decomposition to write f(s) as f(s) = A/(s+3i) + B/(s-3i), where A and B are constants. Solving for A and B, we get A = -2/(3i+2) and B = 2/(3i-2).

Next, we use the table of Laplace transforms to find the inverse Laplace transforms of A/(s+3i) and B/(s-3i). We have L^-1[A/(s+3i)] = -2e^(-3t) sin(6t) u(t) and L^-1[B/(s-3i)] = 2e^(3t) sin(6t) u(t). Thus, the inverse Laplace transform of f(s) is f(t) = L^-1[f(s)] = L^-1[A/(s+3i)] + L^-1[B/(s-3i)] = -2e^(-3t) sin(6t) u(t) + 2e^(3t) sin(6t) u(t) = -2e^(-2t) sin(6t) u(t).

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If[tex]v(X+3) = v(X+6)[/tex], then X has a constant variance.

If we have a random variable X, then we know that:

[tex]v(X) = E[(X - μ)^2],[/tex]

where E is the expectation operator and μ is the mean of X.

Using this formula, we can expand v(X+3) and v(X+6) as follows:

[tex]v(X+3) = E[(X+3 - μ)^2]\\v(X+6) = E[(X+6 - μ)^2][/tex]

Now we can simplify these expressions:

[tex]v(X+3) = E[(X - μ + 3)^2]\\= E[(X - μ)^2 + 6(X - μ) + 9]\\= v(X) + 6E[X - μ] + 9v(X+6) \\= E[(X - μ + 6)^2]\\= E[(X - μ)^2 + 12(X - μ) + 36]\\= v(X) + 12E[X - μ] + 36[/tex]

Since v(X+3) = v(X+6), we can equate the two expressions:

[tex]v(X) + 6E[X - μ] + 9 = v(X) + 12E[X - μ] + 36\\[/tex]

Simplifying this equation yields:

[tex]6E[X - μ] = 27\\E[X - μ] = 4.5[/tex]

Since the expected value of X minus its mean is 4.5, we can say that the mean of X+3 is 4.5 greater than the mean of X. Similarly, the mean of X+6 is 4.5 greater than the mean of X.

Since the variance is a measure of how spread out the data is from its mean, and the difference in the means of X+3 and X+6 is constant, it follows that the variances of X+3 and X+6 must also be the same.

Therefore, we can conclude that if[tex]v(X+3) = v(X+6),[/tex] then X has a constant variance.

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A phlebotomist conducted a study involving 25,525 individuals aged between 35,535 and 44,444 years old to measure their cholesterol levels.

Summary statistics for the sample were obtained to analyze the data and draw conclusions about cholesterol levels in this specific age group. Based on the information given, we know that a phlebotomist measured the cholesterol levels of a sample of 252525 people between the ages of 353535 and 444444 years old. Here are the summary statistics for the samples:

- Sample size: 252525
- Age range: 353535 to 444444 years old
- Cholesterol levels: No information was provided about the mean, median, mode, or range of cholesterol levels in the sample.

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1. The general solution will be y(x) = Ce^(2x), where C is a constant. Now, apply the initial condition (4,5): 5 = Ce^(8). Solving for C, we get C = 5/e^8. So the particular solution is y(x) = (5/e^8)e^(2x).

2. For the general solution of y'(x) = x(x + 12), first integrate both sides of the equation with respect to x to obtain the antiderivative. This gives y(x) = (1/3)x^3 + 6x^2 + C,

3. To find the particular solution using the initial conditions (1,3).Therefore, the particular solution is y(x) = (1/3)x^3 + 6x^2 - 20/3.

For the first question, we need to use the method of integrating factors to find the particular solution. The integrating factor is e^(∫(x-2) dx) = e^(x^2/2 - 2x), which we can use to rewrite the differential equation as (e^(x^2/2 - 2x) y)' = e^(x^2/2 - 2x) (x-2). Integrating both sides with respect to x, we get e^(x^2/2 - 2x) y = ∫e^(x^2/2 - 2x) (x-2) dx. Evaluating the integral, we get e^(x^2/2 - 2x) y = -1/2 e^(x^2/2 - 2x) (x-2)^2 + C, where C is a constant of integration. Plugging in the initial condition (4,5), we can solve for C to get the particular solution y = -1/2 (x-2)^2 + 5.

For the second question, we can use the method of separation of variables to find the general solution. Separating the variables and integrating, we get ∫(1/y) dy = ∫(x+12) dx, which simplifies to ln|y| = (1/2)x^2 + 12x + C, where C is a constant of integration. Exponentiating both sides, we get |y| = e^(1/2 x^2 + 12x + C), which can be rewritten as y = ±e^(1/2 x^2 + 12x + C). Therefore, the general solution is y = C1 e^(1/2 x^2 + 12x) + C2 e^(-1/2 x^2 - 12x), where C1 and C2 are constants of integration.

For the third question, we can use the same method as the first question, but with a different integrating factor. The integrating factor is e^(∫(1-**) dx) = e^(x - **x^2/2), which we can use to rewrite the differential equation as (e^(x - **x^2/2) y)' = e^(x - **x^2/2) (1-**). Integrating both sides with respect to x, we get e^(x - **x^2/2) y = ∫e^(x - **x^2/2) (1-**) dx. Evaluating the integral, we get e^(x - **x^2/2) y = (1-**/2) e^(x - **x^2/2) + C, where C is a constant of integration. Plugging in the initial condition (1,3), we can solve for C to get the particular solution y = (1-**/2) e^(x - **x^2/2) + **/2 + 2.


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The total cost of an appointment that lasts for 3 1/2 hours is 290

From the question, we have the following parameters that can be used in our computation:

y = 60x + 80

Given that

x = 3 1/2

Substitute the known values in the above equation, so, we have the following representation

y = 60 * (3 1/2) + 80

Evaluate the expression

y = 290

Hence, the total cost is 290

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Let's start by finding the length and width of the rectangular field. Let's call the width of the rectangular field "w" and the length "l."

We know that the perimeter (P) of a rectangle is given by:

P = 2l + 2w

And we know that the perimeter of this particular rectangular field is 116 yards. So we can write:

116 = 2l + 2w

We also know that one side of the rectangular field measures 27 yards. Let's assume that this is the length of the field (l), so we can write:

l = 27

Substituting this value into the equation for the perimeter, we get:

116 = 2(27) + 2w

Simplifying:

116 = 54 + 2w

2w = 62

w = 31

So the width of the rectangular field is 31 yards.

The area (A) of a rectangle is given by:

A = l*w

So the area of the rectangular field is:

A = 27*31 = 837

Now we need to find the sides of a square field with the same area as the rectangular field. The area of a square (A_s) is given by:

A_s = s^2

Where s is the length of one side of the square. We know that the area of the square is the same as the area of the rectangular field, so we can write:

A_s = A

s^2 = 837

s = sqrt(837) ≈ 28.948

Rounded to three decimals, the sides of the square field measure approximately 28.948 yards.

The sides of the square field measure approximately 29 yards.

Let the other side of the rectangular field be denoted by x. Since the perimeter of the rectangular field is 116 yards, we have:

2x + 2(27) = 116

2x + 54 = 116

2x = 62

x = 31

So the rectangular field has sides of length 27 and 31, and its area is:

A = 27 × 31 = 837

The area of the square field with the same area as the rectangular field is also 837, so its side length s satisfies:

s^2 = 837

Taking the square root of both sides, we get:

s ≈ 28.997

Rounding to three decimal places, we get s ≈ 29.

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The radius of circle F is equal to: C. 12.

In Mathematics and Geometry, Pythagorean's theorem is modeled or represented by the following mathematical equation (formula):

a² + b² = c²

Where:

a, b, and c represents the length of sides or side lengths of any right-angled triangle.

By substituting the given side lengths into the formula for Pythagorean's theorem, we have the following;

a² + b² = c²

a² + 9² = 15²

a² = 225 - 81

a = √144

a = 12 units.

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The area to the right of 0.75 (i.e., the probability that z is greater than 0.75) is: 1 - 0.7734 = 0.2266 This means that the probability of getting a value greater than 4.5 in this non-standard normal distribution is approximately 0.2266.

To answer this question, we need to use the normal distribution table. However, since the distribution is non-standard, we need to standardize the value of 4.5 using the formula: z = (x - μ) / σ where μ is the mean and σ is the standard deviation.

Substituting the given values, we get: z = (4.5 - 3) / 2 z = 0.75 Now, we need to find the probability that z is greater than 0.75. Looking at the normal distribution table, we find that the area to the left of 0.75 is 0.7734.

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Answer:

C

Step-by-step explanation:

The empirical rule states that approximately 68% of the data falls within one standard deviation, we can estimate that the percentage of lightbulb replacement requests numbering between 44 and 58 is slightly more than half of 68%, which is approximately 34%.

Using the empirical rule, we know that approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations. Since we are looking for the percentage of requests between 44 and 58, which is one standard deviation below and above the mean, we can estimate that approximately 68% / 2 = 34% of the requests fall within this range. Therefore, the approximate percentage of lightbulb replacement requests numbering between 44 and 58 is 34. Based on the given information, the number of daily requests for lightbulb replacements has a mean of 58 and a standard deviation of 7. Using the empirical rule, we know that approximately 68% of the data falls within one standard deviation of the mean. Since the distribution is bell-shaped, we can apply the empirical rule here.In this case, one standard deviation below the mean is 58 - 7 = 51. The requested range is between 44 and 58, which is slightly larger than the range of one standard deviation below the mean (51-58).

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The height of the can in inches is,

⇒ h = 6.37 inches

Since, We know that;

Volume of the can ground coffee = πr²h

where,

r = radius

h = height of the cylinder

Therefore, We get;

r = 1/2 inches

V = 5 cubic inches

Hence, We get;

Volume of the can ground coffee = 3.14 × (1/2)² × h

5 = 3.14 × 1/4 × h

20/3.14 = h

h = 6.37 inches

Hence, The height of the can in inches is,

⇒ h = 6.37 inches

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To find the average mass of a paper clip in the given sample, we need to divide the total mass of the paper clips (10.135 g) by the number of paper clips (10). Therefore, the average mass of a paper clip in this sample is 1.0135 g (10.135 g / 10 paper clips).

The average mass of a paper clip in this sample, you need to follow these steps:
1. Determine the total mass of the paper clips: 10.135 g
2. Count the number of paper clips in the sample: 10
3. Calculate the average mass by dividing the total mass by the number of paper clips:
Average mass = Total mass / Number of paper clips
Average mass = 10.135 g / 10
Average mass ≈ 1.0135 g
The average mass of a paper clip in this sample is approximately 1.0135 g.

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It will take approximately 18.42 years for the price of beef to triple with a 2% inflation compounded biannually.

To find out how long it will take for the price of beef to triple with a 2% inflation compounded biannually, we'll use the compound interest formula:

Final amount = Initial amount * (1 + interest rate)^number of periods

Here, we want the final amount to be triple the initial amount, so we have:

3 * Initial amount = Initial amount * (1 + 0.02)^number of periods

Divide both sides by the Initial amount:

3 = (1 + 0.02)^number of periods

Now, we need to solve for the number of periods. To do this, we'll use the logarithm:

log(3) = log((1 + 0.02)^number of periods)

Using the logarithm property log(a^b) = b*log(a), we get:

log(3) = number of periods * log(1.02)

Now, we'll solve for the number of periods:

number of periods = log(3) / log(1.02) ≈ 36.84

Since the inflation is compounded biannually, we need to divide the number of periods by 2 to get the number of years:

number of years = 36.84 / 2 ≈ 18.42

So, it will take approximately 18.42 years for the price of beef to triple with a 2% inflation compounded biannually.

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a. The probability that three people have different birthdates is 0.9918.

b. The probability that ten people have different birthdates is 0.883

c. The effect of birth rate on the probability of 10 randomly chosen people having different birthdates depends on the specifics of the distribution of birthdates.

a. To find the probability that three randomly chosen people have different birthdates, we can use the principle of inclusion-exclusion.

The probability that the first person has a unique birthdate is 1.

The probability that the second person has a birthdate different from the first is 364/365, and the probability that the third person has a birthdate different from the first two is 363/365.

Therefore, the probability that three people have different birthdates is:

1 x 364/365 x 363/365 = 0.9918

b. To find the probability that 10 randomly chosen people all have different birthdates, we can similarly use the principle of inclusion-exclusion.

The probability that the first person has a unique birthdate is 1.

The probability that the second person has a birthdate different from the first is 364/365, and the probability that the third person has a birthdate different from the first two is 363/365.

Continuing in this way, the probability that the tenth person has a birthdate different from the first nine is 356/365.

Therefore, the probability that ten people have different birthdates is:

1 x 364/365 x 363/365 x ... x 356/365 = 0.883

c. If birth rates are higher during some parts of the year than others, then it is possible that there will be more people with the same birthdate, which would decrease the probability that 10 randomly chosen people have different birthdates.

However, the effect of this depends on the distribution of birthdates.

For example, if there are more people born in the summer months, then it may be more likely that the 10 randomly chosen people have different birthdates, since they are less likely to share a birthdate in common.

On the other hand, if there are more people born in a specific month, then it may be more likely that the 10 randomly chosen people share a birthdate in common.

Ultimately, the effect of birth rate on the probability of 10 randomly chosen people having different birthdates depends on the specifics of the distribution of birthdates.

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To determine the probability that a patient who tests positive for malignalitaloptereosis actually has the disease, we need to use the positive predictive value (PPV) formula:  PPV = true positives / (true positives + false positives)

In this case, we are given the sensitivity and specificity of the test, but we do not have information about the prevalence of the disease in the population being tested. Without knowing the prevalence, we cannot calculate the true positives or false positives and therefore cannot calculate the PPV. However, the information given only provides the sensitivity (0.92) and specificity (0.77) of the test. To calculate the PPV, we also need the prevalence of the disease in the population. Since the prevalence is not provided, there is not enough information given to determine the probability that a patient with a positive test result actually has the disease. Therefore, the correct answer is: Not enough information given.

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If a sector of a circle has a central angle measure of 90°, and an area of 7 square inches then the area of the entire circle is 35.44 square inches.

We can use the formula for the area of a sector to solve this problem:

Area of sector = (θ/360)πr²

We are given that the central angle measure is 90°,

so θ = 90.

Area of the sector is 7 square inches.

We can set up an equation:

7 = (90/360)πr²

7 = (1/4)πr²

Multiplying both sides by 4/π, we get:

28/π = r²

r =3.36 inches

Now that we know the radius of the circle, we can find its area using the formula:

Area of circle = πr²

Substituting r = 3.36, we get:

Area of circle = 35.44 square inches

Therefore, the area of the entire circle is 35.44 square inches.

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The rate of change of the distance from the particle to the origin at this instant is 12.247 units/second.

Let's call the distance from the particle to the origin at a certain point (x, y) as d(x, y). Then, by the Pythagorean theorem, we have:

d(x, y) = √(x^2 + y^2)

We want to find the rate of change of d(x, y) with respect to time t, which we can write as:

d/dt [d(x, y)]

To find this, we need to express d(x, y) in terms of t. We know that the particle is moving along the curve y = 2√5x + 11, so we can substitute this into the equation for d(x, y):

d(x, y) = √(x^2 + y^2) = √(x^2 + (2√5x + 11)^2)

Now we can use the chain rule to find d/dt [d(x, y)]:

d/dt [d(x, y)] = d/dt [√(x^2 + (2√5x + 11)^2)]

= (1/2) (x^2 + (2√5x + 11)^2)^(-1/2) * d/dt [x^2 + (2√5x + 11)^2]

We already know that dx/dt = 5, so we just need to find dy/dt:

dy/dx = d/dx [2√5x + 11] = √5

dy/dt = dy/dx * dx/dt = √5 * 5 = 5√5

Now we can substitute dx/dt and dy/dt into the expression we found for d/dt [d(x, y)]:

d/dt [d(x, y)] = (1/2) (x^2 + (2√5x + 11)^2)^(-1/2) * (2x + 4(2√5x + 11) dx/dt)

= (1/2d(x, y)) (x + 4(√5x + 11)) dx/dt

Finally, we can substitute the values for x and dx/dt that we know from the problem:

x = 5

dx/dt = 5

And we can substitute the expression we found for d(x, y) back into the equation for d/dt [d(x, y)]:

d/dt [d(x, y)] = (1/2√(x^2 + (2√5x + 11)^2)) (x + 4(√5x + 11)) dx/dt

= (1/2√(5^2 + (2√5(5) + 11)^2)) (5 + 4(√5(5) + 11)) (5)

Simplifying this expression gives:

d/dt [d(x, y)] ≈ 12.247 units/second

So the rate of change of the distance from the particle to the origin at the instant when the particle passes through the point (5, 12) is approximately 12.247 units/second.

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One way to solve this system of equations is to use the elimination method. You can solve for one variable in terms of the other using one equation and then substitute that expression into the other equation. For example, you can solve for x in terms of y using the second equation: x = -11 - y. Then substitute this expression for x into the first equation: (-11 - y) * y = -80. Solving this quadratic equation will give you the values of y, and then you can find the corresponding values of x.

The first derivative of the function f(x) = x^4 - 128.22x^2 - 1 is f'(x) = 4x^3 - 2(128.22)x, and the second derivative is f''(x) = 12x^2 - 2(128.22). The intervals are (-∞, 5) U (7, ∞).

Let's find the derivatives and discuss the intervals.

To calculate the first derivative, f'(x), we apply the power rule: f'(x) = 4x^3 - 2(128.22)x. Now let's find the second derivative, f''(x). Using the power rule again, we get f''(x) = 12x^2 - 2(128.22).

The intervals you mentioned are (-∞, 5) U (7, ∞). These represent two separate infinite intervals, the first one ranging from negative infinity to 5 and the second ranging from 7 to positive infinity.

In summary, the first derivative of the function f(x) = x^4 - 128.22x^2 - 1 is f'(x) = 4x^3 - 2(128.22)x, and the second derivative is f''(x) = 12x^2 - 2(128.22). The intervals are (-∞, 5) U (7, ∞).

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The answer is g(1, 2) = -2, g(2, 1) = 7, g(1, 1) = 1, g(-1, 1) = 1, and g(2, -1) = 17.

The given function is g(x, y) = 2x^2 - y^2.

Substituting x = 1 and y = 2, we get g(1, 2) = 2(1)^2 - (2)^2 = -2.

Substituting x = 2 and y = 1, we get g(2, 1) = 2(2)^2 - (1)^2 = 7.

Substituting x = 1 and y = 1, we get g(1, 1) = 2(1)^2 - (1)^2 = 1.

Substituting x = -1 and y = 1, we get g(-1, 1) = 2(-1)^2 - (1)^2 = 1.

Substituting x = 2 and y = -1, we get g(2, -1) = 2(2)^2 - (-1)^2 = 17.

Therefore, g(1, 2) = -2, g(2, 1) = 7, g(1, 1) = 1, g(-1, 1) = 1, and g(2, -1) = 17.

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The solution using the Laplace transform is f(t) = 3t^2 - e^-t + ∫0^t f(τ)e^(τ-t)dτ.

To solve for f(t) in the equation f(t) = 3t^2 - e^-t - f(τ)e^(t-τ)dτ, we need to use the Laplace transform. We will apply the Laplace transform on both sides of the equation, and then solve for F(s), where F(s) is the Laplace transform of f(t).

Applying the Laplace transform on both sides of the equation, we get F(s) = 3(2/s^3) - (1/(s+1)) - F(s)E(s), where E(s) is the Laplace transform of e^(t-τ)dτ.

We can simplify this expression to solve for F(s):

F(s) + F(s)E(s) = 6/s^3 - 1/(s+1)

F(s) (1 + E(s)) = 6/s^3 - 1/(s+1)

F(s) = (6/s^3 - 1/(s+1)) / (1 + E(s))

Finally, we need to find the inverse Laplace transform of F(s) to get the solution for f(t). This can be done using partial fractions and the inverse Laplace transform tables.

Hence  the solution is f(t) = 3t^2 - e^-t + ∫0^t f(τ)e^(τ-t)dτ.

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The expression in terms of I, E, and Y will be E / (I - Y).

Given that:

Equation, I = E/X + Y

The definition of simplicity is making something simpler to achieve or grasp while also making it a little less difficult.

Simplify the equation for X, then we have

I = E/X + Y

I - Y = E/X

X = E / (I - Y)

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(a) The probability that all 4 missiles will fire is 0.3297.

(b) The probability that at most 2 missiles will not fire is 0.9576.

(a) The probability that all 4 missiles will fire can be calculated as follows:

The total number of ways to select 4 missiles from a lot of 14 missiles is given by the combination formula: C(14, 4) = 1001.

The number of ways to select 4 non-defective missiles from a lot of 11 non-defective missiles is given by the combination formula: C(11, 4) = 330.

Therefore, the probability that all 4 missiles will fire is:

P(all 4 fire) = C(11, 4) / C(14, 4) = 330 / 1001 = 0.3297 (rounded to four decimal places)

(b) The probability that at most 2 missiles will not fire can be calculated as follows:

The number of ways to select 4 missiles from a lot of 14 missiles that contain 3 defective missiles is given by the combination formula: C(11, 4) * C(3, 0) + C(11, 3) * C(3, 1) + C(11, 2) * C(3, 2) = 6084.

The number of ways to select 4 missiles from a lot of 14 missiles that do not contain any defective missiles is given by the combination formula: C(11, 4) * C(3, 0) = 330.

Therefore, the probability that at most 2 missiles will not fire is:

P(at most 2 do not fire) = [C(11, 4) * C(3, 0) + C(11, 3) * C(3, 1) + C(11, 2) * C(3, 2)] / C(14, 4) + C(11, 4) * C(3, 0) / C(14, 4) = 0.9576 (rounded to four decimal places)

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The planes are approximately 1031 miles from each other 2 hours after takeoff. To solve this problem, we need to use trigonometry to find the distance between the two planes after 2 hours of flight. We can use the law of cosines to find the distance between the two planes.

First, we need to find the distances each plane has traveled after 2 hours. The plane traveling on a bearing S 13° W has traveled:

d1 = 450 miles/hour * 2 hours = 900 miles

The plane traveling on a bearing N 75° E has traveled:

d2 = 250 miles/hour * 2 hours = 500 miles

Now we can use the law of cosines to find the distance between the two planes:

d^2 = d1^2 + d2^2 - 2*d1*d2*cos(103°)

d^2 = 900^2 + 500^2 - 2*900*500*cos(103°)

d^2 = 810,000 + 250,000 - 900,000*(-0.287)

d^2 = 1,060,433

d = 1030.9 miles

Therefore, the planes are approximately 1031 miles from each other 2 hours after takeoff.

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To find the number of ways to pick a set of six balls from the bin in which at least one ball has an odd number, we can use the principle of inclusion-exclusion.

First, we find the total number of ways to pick a set of six balls from the bin, which is 21 choose 6 (written as C(21,6)) = 54264.

Next, we find the number of ways to pick a set of six balls from the bin in which all the balls have even numbers. There are only 10 even-numbered balls in the bin, so the number of ways to pick a set of six even-numbered balls is 10 choose 6 (written as C(10,6)) = 210.

Therefore, the number of ways to pick a set of six balls from the bin in which at least one ball has an odd number is:

C(21,6) - C(10,6) = 54264 - 210 = 54054.

So there are 54054 ways to pick a set of six balls from the bin in which at least one ball has an odd number.e

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The center of the circle is (-4, 3) and the radius is 3.

In order to calculate the center, (h, k), and radius, r, of the circle outlined by the equation x² + y² + 8x - 6y + 16 = 0, we can reposition the given equation into its most conventional shape for a circle:

(x - h)² + (y - k)² = r².

Completing the square by adding and subtracting (8/2)² on account of x and (6/2)² due to y,

(x² + 8x + 16) + (y² - 6y + 9) = -16 + 16 + 9

(x + 4)² + (y - 3)² = 9

This transformation thus showcases that the center of the circle is (-4, 3), as (h, k) equals (-4, 3). Additionally, the radius is the square root of r², or √9 = 3.

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A conical water tank with vertex down has a radius of 12 feet at the top and is 27 feet high. If water flows into the tank at a rate of 10 cubic feet per minute, the rate at which the water level rises depends on the volume of the cone at a specific moment.

The formula to find the volume of a cone is V = (1/3)πr²h, where V is the volume, r is the radius, and h is the height. Since the dimensions of the cone are given, you can use similar triangles to find the relationship between the radius (r) and the height (h) of the water in the tank at any moment.

Given: r_cone = 12 ft, h_cone = 27 ft, and dV/dt = 10 ft³/min.

Let r_water and h_water represent the radius and height of the water at any given moment. Using similar triangles, we have:

r_water / h_water = r_cone / h_cone
r_water = (12 ft / 27 ft) * h_water

Now, substitute this relationship into the cone volume formula:

V = (1/3)π((12 ft / 27 ft) * h_water)² * h_water

Differentiate this equation with respect to time to find the rate at which the water level rises:

dV/dt = d(π/3 * (144/729) * h_water³)/dt
10 ft³/min = (π/3 * (144/729)) * 3 * h_water² * dh_water/dt

Solve for dh_water/dt:

dh_water/dt = 10 ft³/min / (π * (144/729) * h_water²)

This equation shows the rate at which the water level rises (dh_water/dt) at any given moment based on the current height of the water (h_water) in the tank.

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