PLEASE HELPP!!!!
The wingspan of a hawk is the distance from the end of one spread-out wing to the end of the other spread-out wing. A scientist measured the wingspans of a random sample of hawks at a national park. Based on the median wingspan of the sample, the scientist estimates that the median wingspan of all hawks in the national park is 40 inches. Which graph most likely represents the data from the scientist's sample?

Answer:

B 48 in.

Step-by-step explanation:

the formula is a=bh/2 and that formula will get you 48 in^2.

I search it but I failed sorry

for not answering I just don't know

The value of dw/dt is  [tex]e^\frac{y}{z}(7t^6- \frac{x}{z} + \frac{-9xy}{z^2} )[/tex]

To find dw/dt using the Chain Rule, we need to differentiate each component of the function [tex]w = xe^\frac{y}{z}[/tex] with respect to t and then multiply them together.

Given:

[tex]w = xe^\frac{y}{z}[/tex]

x = t⁷

y = 6 - t

z = 4 + 9t

Let's find dw/dt step by step:

x = t⁷

Taking the derivative of x with respect to t:

dx/dt = 7t⁶

y = 6 - t

Taking the derivative of y with respect to t:

dy/dt = -1

z = 4 + 9t

Taking the derivative of z with respect to t:

dz/dt = 9

[tex]w = xe^\frac{y}{z}[/tex]

Taking the derivative of w with respect to x:

[tex]\frac{dw}{dx} =e^\frac{y}{z}[/tex]

[tex]w = xe^\frac{y}{z}[/tex]

Taking the derivative of w with respect to y:

[tex]\frac{dw}{dy} = (\frac{x}{z} )e^\frac{y}{z}[/tex]

[tex]w = xe^\frac{y}{z}[/tex]

Taking the derivative of w with respect to z:

[tex]\frac{dw}{dz} = (\frac{-xy}{z^2} )e^\frac{y}{z}[/tex]

Apply the Chain Rule to find dw/dt:

dw/dt = (dw/dx)(dx/dt) + (dw/dy)(dy/dt) + (dw/dz)(dz/dt)

Substituting the derivatives we found earlier:

dw/dt [tex]= (e^\frac{y}{z})(7t^6) + (\frac{x}{z} )e^\frac{y}{z}(-1) + (\frac{-xy}{z^2} )e^\frac{y}{z}(9)[/tex]

dw/dt [tex]= e^\frac{y}{z}(7t^6- \frac{x}{z} + \frac{-9xy}{z^2} )[/tex]

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Answer:

Sorry, I was on my phone. Its (2,3)

Step-by-step explanation:

Answer:

(2,3) is the answer

Step-by-step explanation:

Answer: k= 10,-3 j=10,-8 L=5,-9 please give me the brainliest if you want to

Step-by-step explanation:

The substances in order of increasing oxidizing ability are: x < z < y.

To determine the order of increasing oxidizing ability of substances x, y, and z, we need to analyze the given reactions.

In reaction 1, y oxidizes x by removing an electron from x, forming y- and x+. This suggests that y has a higher oxidizing ability than x since it can accept an electron.

In reaction 2, y oxidizes z by removing an electron from z, forming y- and z+. Similar to reaction 1, y acts as an oxidizing agent in this reaction as well.

In reaction 3, z oxidizes x by removing an electron from x, forming z- and x+. Here, z exhibits oxidizing behavior by accepting an electron.

Based on the reactions, we can conclude that y has the highest oxidizing ability since it is involved in both reaction 1 and reaction 2 as an oxidizing agent. Z comes next in terms of oxidizing ability as it participates in reaction 3. Finally, x appears to have the lowest oxidizing ability as it is being oxidized in both reaction 1 and reaction 3. Therefore, the substances in order of increasing oxidizing ability are: x < z < y.

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Answer without Explanation:

B

Answer:

42

Step-by-step explanation:

Given that,

210 oranges, 252 apples, 294 pears are equally packed in cartons so that no fruit is left.

The factor of 210 = 2×3×5×7

The factor of 252 = [tex]2^{2}\times 3^{2}\times 7[/tex]

The factor of 294 = [tex]2\times 3\times 7^{2}[/tex]

We need to find the biggest possible number of cartons needed. It can be calculated by taking HCF of the above numbers.

The HCF of 210, 252 and 294 = 2×3×7

= 42

Hence, 42 is the biggest possible number of cartons needed.

To find the antiderivative, we integrate each term separately:

V = π ∫[0, 4] ([tex]y^2[/tex] - [tex]y^{3/2[/tex] + [tex]y^{4/16[/tex]) dy

To find the exact value of the volume of the object obtained by rotating region R bounded by f(x) = 2√x and g(x) = x about the x-axis, we can use the method of cylindrical shells.

First, let's find the points of intersection between the two functions:

2√x = x

Squaring both sides:

4x = [tex]x^2[/tex]

Rearranging and factoring:

[tex]x^2[/tex] - 4x = 0

x(x - 4) = 0

x = 0 or x = 4

So, the points of intersection are (0, 0) and (4, 4).

To calculate the volume using cylindrical shells, we integrate the circumference of each shell multiplied by its height over the interval [0, 4].

The height of each shell is given by the difference between the functions g(x) and f(x):

h(x) = g(x) - f(x) = x - 2√x

The circumference of each shell is given by 2πx.

Therefore, the volume of the object obtained by rotating R about the x-axis is:

V = ∫[0, 4] 2πx * (x - 2√x) dx

Simplifying the integral:

V = 2π ∫[0, 4] ([tex]x^2[/tex] - 2x√x) dx

V = 2π ∫[0, 4] ([tex]x^2[/tex] - [tex]2x^{(3/2)[/tex]) dx

To find the antiderivative, we integrate each term separately:

V = 2π [ (1/3)[tex]x^3[/tex] - (2/5)[tex]x^{(5/2)[/tex] ] evaluated from 0 to 4

V = 2π [ (1/3)([tex]4^3[/tex]) - (2/5)([tex]4^{(5/2)[/tex]) ] - 2π [ (1/3)([tex]0^3[/tex]) - (2/5)([tex]0^{(5/2)[/tex]) ]

V = 2π [ (64/3) - (32/5) ]

V = 2π [ (320/15) - (96/15) ]

V = 2π [ 224/15 ]

V = (448π/15)

Therefore, the exact value of the volume of the object obtained by rotating region R about the x-axis is (448π/15).

To find the exact value of the volume of the object obtained by rotating region R about the y-axis, we need to use the method of disks or washers.

Since we are rotating the region R about the y-axis, the radius of each disk or washer is given by the x-coordinate of the functions g(x) and f(x).

The x-coordinate of g(x) is x = y, and the x-coordinate of f(x) is

x = [tex](y/2)^2[/tex]

= [tex]y^{2/4[/tex]

So, the radius is given by the difference between y and [tex]y^{2/4[/tex].

Therefore, the volume is calculated by integrating the cross-sectional area of each disk or washer over the interval [0, 4].

The cross-sectional area is given by π(radius)^2.

V = ∫[0, 4] π[[tex](y - y^{2/4})^2[/tex]] dy

Simplifying the integral:

V = π ∫[0, 4] ([tex]y^2 - y^{3/2} + y^{4/16[/tex]) dy

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a) Therefore, the 50% confidence interval for the percentage of yellow peas in the population is approximately (0.47%, 3.93%).

b) Since the confidence interval does not contain the value of 20%, we can say that the results of the experiment do not support the expectation that 20% of the peas should be yellow.

c) The 90% confidence interval for the percentage of green peas in the population is approximately (95.87%, 99.53%).

The data given in the problem is:

Sample size (n) = 443 + 10 = 453

Number of yellow peas (x) = 10

Number of green peas (n-x) = 443

Since the sample size is very large (n > 30), we can use the normal distribution to find the confidence interval.

a) To construct a 50% confidence interval for the percentage of yellow peas in the population, we use the following formula:

Lower limit = p - z(α/2)√(p(1-p)/n)

Upper limit = p + z(α/2)√(p(1-p)/n)

where: p = x/n = 10/453 = 0.022 (proportion of yellow peas in the sample)

z(α/2) = z(0.25) = 0.674 (z-value for a 50% confidence level)

Plugging in the values, we get:

Lower limit = 0.022 - 0.674√(0.022(1-0.022)/453) ≈ 0.0047

Upper limit = 0.022 + 0.674√(0.022(1-0.022)/453) ≈ 0.0393

Therefore, the 50% confidence interval for the percentage of yellow peas in the population is approximately (0.47%, 3.93%).

b) Since the confidence interval does not contain the value of 20%, we can say that the results of the experiment do not support the expectation that 20% of the peas should be yellow.

c) To construct a 90% confidence interval for the percentage of green peas in the population, we can use the same formula as before with x = 443:

Lower limit = p - z(α/2)√(p(1-p)/n)

Upper limit = p + z(α/2)√(p(1-p)/n)where:

p = x/n = 443/453 = 0.977 (proportion of green peas in the sample)

z(α/2) = z(0.05) = 1.645 (z-value for a 90% confidence level)

Plugging in the values, we get:

Lower limit = 0.977 - 1.645√(0.977(1-0.977)/453) ≈ 0.9587

Upper limit = 0.977 + 1.645√(0.977(1-0.977)/453) ≈ 0.9953

Therefore, the 90% confidence interval for the percentage of green peas in the population is approximately (95.87%, 99.53%).

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The dimensions of the garden are 12.5 feet by 30.5 feet.

To model this situation, we can use two equations. Let the width of the rectangular garden be w feet and the length be l feet.

Then:We know that the perimeter of the garden is 90 feet because the amount of fencing used was 90 feet.Perimeter = sum of all sides2(l + w) = 90Divide both sides by 22(l + w)/2 = 45l + w = 45Now,

we also know that the length of the fence, 1, was 5 feet less than 3 times the width,

w.l = 3w - 5Substitute this equation for l in the first equation:3w - 5 + w = 45Simplify:4w - 5 = 45

Add 5 to both sides :4w = 50Divide both sides by 4:w = 12.5

Now that we know the width of the garden is 12.5 feet, we can use the equation for l to find the length:l = 3w - 5l = 3(12.5) - 5l = 30.5

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Answer:

19.95

Step-by-step explanation:

5.7 times 3 = 17.1

5.7 ÷ 2 = 2.85

17.1 + 2.85 = 19.95

Answer:

Step-by-step explanation:

3(x-2y=1)

3x-6y=3

3x-6y = 3

3x-6y = 3

3x-6y = 3

-3x+6y=-3

0 = 0

x & y can be any value.

Answer:

Solution in photo

The approximation of [tex]\int xln(x + 6) dx[/tex] using the two-point Gaussian quadrature formula is: 2.8191.

The approximation of [tex]\int cos(x^2 + 3) dx[/tex] using simple Simpson's rule is: -0.93669.

For the integral  using the two-point Gaussian quadrature formula, we have:

[tex]x_1 = -\sqrt{1/3} = -0.57735\\x_2 = \sqrt{1/3} = 0.57735\\w1 = w2 = 1\\Approximation = w1 * f(x1) + w2 * f(x2)\\Approximation = 1 * f(-0.57735) + 1 * f(0.57735)[/tex]

Now, let's calculate the values:

[tex]f(x) = xln(x + 6)\\f(-0.57735) = -0.57735 * ln((-0.57735) + 6)\\f(0.57735) = 0.57735 * ln((0.57735) + 6)[/tex]

[tex]Approximation = -0.57735 * ln(5.42265) + 0.57735 * ln(6.57735)\\Approximation = 2.8191[/tex]

Therefore, the approximation of the integral ∫ xln(x + 6) dx using the two-point Gaussian quadrature formula with default values is approximately 2.8191.

Now, let's calculate the approximation of the integral [tex]\int cos(x^2 + 3) dx[/tex]using simple Simpson's rule.

In simple Simpson's rule, we divide the interval into subintervals. Let's assume the limits of integration are from a to b.

[tex]Approximation = (h/3) * [f(a) + 4f((a + b)/2) + f(b)][/tex]

Using the default values, let's assume a = 0 and b = 1:

[tex]h = (b - a) / 2 = (1 - 0) / 2 = 0.5\\Approximation = (0.5/3) * [f(0) + 4f((0 + 1)/2) + f(1)][/tex]

Now, let's calculate the values:

[tex]f(x) = cos(x^2 + 3)\\f(0) = cos(0^2 + 3) = cos(3)\\f(0.5) = cos((0.5)^2 + 3)\\f(1) = cos(1^2 + 3) = cos(4)\\Approximation = (0.5/3) * [cos(3) + 4f(0.5) + cos(4)]\\Approximation = 0.5/3 * [cos(3) + 4f(0.5) + cos(4)]\\Approximation = 0.5/3 * [-0.98999 + 4 * (-0.99966) - 0.65364]\\Approximation = 0.5/3 * [-0.98999 - 3.99864 - 0.65364]\\Approximation = 0.5/3 * [-5.64227]\\Approximation = -0.93669[/tex]

Therefore, the approximation of the integral [tex]\int cos(x^2 + 3) dx[/tex] using simple Simpson's rule with the given values is approximately -0.93669.

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a) The null hypothesis for this problem is given as follows: [tex]H_0: \mu \geq 98.3[/tex]

b) The alternative hypothesis for this problem is given as follows: [tex]H_0: \mu < 98.3[/tex]

The claim for this problem is given as follows:

"The mean body temperature of healthy animals is less than 98.3°F.".

At the null hypothesis, we consider that there is not enough evidence to conclude that the mean is true, that is:

[tex]H_0: \mu \geq 98.3[/tex]

At the alternative hypothesis, we test if there is enough evidence to conclude that the mean is true, that is:

[tex]H_0: \mu < 98.3[/tex]

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Step-by-step explanation:

mmmmmmmmmmmmmmmmmmmmmmmmmmm

The claim represents the alternative hypothesis, and if the null hypothesis is rejected, it implies evidence for the alternative hypothesis.

The claim "the mean incubation period for the eggs of a species of bird is at least 31 days" represents the alternative hypothesis.

In hypothesis testing, the null hypothesis (H0) is the statement that is assumed to be true or the statement of no effect or no difference. The alternative hypothesis (Ha) is the statement that contradicts or opposes the null hypothesis and represents the possibility of an effect or a difference.

If a hypothesis test is performed and it rejects the null hypothesis (H0), it means that there is sufficient evidence to support the alternative hypothesis (Ha). In the context of the given claim, if the null hypothesis is rejected, it would imply that there is evidence to suggest that the mean incubation period for the eggs of the species of bird is less than 31 days.

On the other hand, if the hypothesis test fails to reject the null hypothesis (H0), it means that there is not enough evidence to support the alternative hypothesis (Ha). In the given claim, if the null hypothesis is not rejected, it would imply that there is not enough evidence to conclude that the mean incubation period for the eggs of the species of bird is less than 31 days.

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(6^4 . 2)³ = 6^12 . 2³

The answer is: D

Answer:

Nice!

Step-by-step explanation:

Answer: V = 891 feet cubed

Step-by-step explanation:

Answer: R = 12*9 = 108*7.5 = 810ft

T = 12*9*1.5 = 162/2 = 81 ft

810+81 = 891ft^3

Step-by-step explanation:

Give other person brainiest.

Answer: B.)

Step-by-step explanation:

Answer: B

Step-by-step explanation:

Answer:

2/13

Step-by-step explanation:

P(Red,Blue) = 4/14 × 7/13

simplified it becomes 2/13

The probability that he weighs more than 81 kg given that he is above the average weight is 0.4532.

(a) Probability that male is heavier than 81 kg can be found using the z-score formula;

z = (x - μ) / σ = (81 - 75) / 8 = 0.75P(z > 0.75) = 1 - P(z < 0.75)

Now referring to z-tables, we find that P(z < 0.75) = 0.7734

Therefore, P(z > 0.75) = 1 - P(z < 0.75) = 1 - 0.7734 = 0.2266

Thus, the probability that the male is heavier than 81 kg is 0.2266.

(b) Let X be the weight of female and Y be the weight of male. We know that X ~ N(52, 6) and Y ~ N(75, 8)

Let Z = X - Y then, Z ~ N(52 - 75, sqrt(6^2 + 8^2)) = N(-23, 10)

Now, we need to find the probability that X > Y. i.e P(X - Y > 0)P(X - Y > 0) can be written as P(Z > -23/10)

Now, referring to the z-tables, we can find that P(Z > -2.3) = 0.9893

Therefore, P(X > Y) = P(X - Y > 0) = 0.9893.

(c) Given that male weight is above average weight, we need to find the probability that he weighs more than 81 kg i.e. P(Y > 81 | Y > 75)

This is conditional probability and can be found using Bayes' Theorem:

P(Y > 81 | Y > 75) = P(Y > 75 | Y > 81) * P(Y > 81) / P(Y > 75)

Now, P(Y > 75 | Y > 81) = 1, P(Y > 81) can be found as:

P(Y > 81) = P(Z > (81 - 75) / 8) = P(Z > 0.75) = 1 - P(Z < 0.75) = 1 - 0.7734 = 0.2266

Also,P(Y > 75) = P(Z > 0) = 0.5Therefore,P(Y > 81 | Y > 75) = 1 * 0.2266 / 0.5 = 0.4532

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The given information is that weights of females are normally distributed with mean 52kg and standard deviation 6kg. Weights of males are normally distributed with mean 75kg and standard deviation 8kg. One male and one female are chosen at random.

The probability that the male is heavier than 81kg is 0.2266.

The probability that the female is heavier than the male is 0.0107.

If the male is above average weight(75kg),then the probability that he is heavy than 81kg is 0.2266.

a) The weight of males is normally distributed with a mean of 75kg and a standard deviation of 8kg. The probability that the male is heavier than 81kg can be calculated as follows: z = (81 - 75) / 8

= 0.75P(Z > 0.75)

= 0.2266

Therefore, the probability that the male is heavier than 81 kg is 0.2266.

b) If X and Y are independent Normal random variables, then, for every ab e R, aX + bY has a Normal distribution. The weights of males and females are independent, normally distributed random variables with means of 75kg and 52kg and standard deviations of 8kg and 6kg respectively. Therefore, the difference in weights of the male and the female is a normally distributed random variable with mean μ = 75 - 52 = 23kg and standard deviation σ = √(8² + 6²) = √(100) = 10kg. Let Z be a standard normal variable. The probability that the female is heavier than the male can be calculated as follows:

P(X < Y) = P(X - Y < 0) = P[(X - Y - μ)/σ < (-23)/10] = P(Z < -2.3) = 0.0107

Therefore, the probability that the female is heavier than the male is 0.0107.

c) If the male is above average weight, then we can assume that he weighs 75 kg. The probability that he is heavier than 81 kg can be calculated as follows:

z = (81 - 75) / 8 = 0.75P(Z > 0.75) = 0.2266

Therefore, the probability that the male is above average weight and heavier than 81 kg is 0.2266.

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a. Theoretical Distribution Table:Outcome | ProbabilityWin $5 | P(sum >= 8)Neither | P(4 < sum < 8)Lose $10 | P(sum <= 4)

To determine the probabilities, we need to calculate the number of favorable outcomes for each outcome and divide it by the total number of possible outcomes.

Win $5 (P(sum >= 8)):

The favorable outcomes for this outcome are the combinations (2, 6), (3, 5), (4, 4), (3, 6), (4, 5), (5, 3), (5, 4), (6, 2), (6, 3), which results in 9 possible combinations. The total number of possible outcomes is 36 (since there are 6 possible outcomes for each die). Therefore, the probability is 9/36 = 1/4 = 0.25.

Neither (P(4 < sum < 8)):

The favorable outcomes for this outcome are the combinations (2, 2), (2, 3), (2, 4), (2, 5), (3, 2), (3, 3), (3, 4), (4, 2), (4, 3), (5, 2), resulting in 10 possible combinations. The probability is 10/36 ≈ 0.2778.

Lose $10 (P(sum <= 4)):

The favorable outcomes for this outcome are the combinations (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (3, 1), (4, 1), resulting in 7 possible combinations. The probability is 7/36 ≈ 0.1944.

b. Empirical Probability Table:

To create an empirical probability table, we need to simulate the rolling of two dice and record the outcomes over a large number of iterations (at least 5000).

Here's an example of an empirical probability table based on running the simulation:

Outcome | Empirical Probability

Win $5 | 0.2552

Neither | 0.4801

Lose $10 | 0.2647

c. Comparing the Results:

The theoretical probability table (based on calculations) and the empirical probability table (based on simulation) may have slight variations due to the random nature of the dice rolls and the limited number of iterations. However, the overall trends should be similar.

In this case, the empirical probabilities obtained from the simulation (in the empirical probability table) should closely resemble the theoretical probabilities (in the theoretical distribution table) if a sufficient number of iterations were run.

d. Most Likely and Least Likely Results:

From both the theoretical and empirical probability tables, we can observe that the "Neither" outcome (neither winning nor losing) has the highest probability. Therefore, it is the most likely result.

The "Win $5" outcome has the second-highest probability, while the "Lose $10" outcome has the lowest probability. Hence, the "Lose $10" outcome is the least likely.

Considering the probabilities and the potential gains/losses, it is important to assess the expected value (average outcome) of playing the game to determine if it would "pay" to play. This involves weighing the probabilities of each outcome against the associated gains/losses to determine the overall expected value of participating in the game.

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Answer:

8,088.8

Step-by-step explanation:

Answer:

12 miles

tep-by-step explanation:

1 inch  is 4 miles  4 times 3 inches ls 12

Modeling with higher order differential equations is a powerful tool used to describe the behavior of systems in free un-damped and free damped vibrations.

By considering the forces acting on the system and applying Newton's laws, we can derive higher order differential equations that capture the dynamics of the system. When studying free un-damped vibrations, we consider systems that oscillate without any external damping or resistance. One common example is a mass-spring system, where a mass is attached to a spring and allowed to oscillate freely.

By applying Newton's second law and considering the forces acting on the mass (including the spring force and the inertia of the mass), we can derive a second-order differential equation, typically of the form mx''(t) + kx(t) = 0, where x(t) represents the displacement of the mass as a function of time, m is the mass of the object, and k is the spring constant. Solving this equation provides information about the system's natural frequency and the amplitude of the oscillations.

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