the p-value method for testing hypotheses is often preferred by statisticians because multiple select question. 1.you do not need to worry about type ii errors. 2.it expresses the strength of your evidence against the null. 3.it does not involve a test statistic. 4.it is more flexible than the critical value method.

equation of line

y=mx+b

If m=-3/2

we have

From points (2,3), we find b

Therefore

According to the Central Limit Theorem, for a sufficiently large sample size, the distribution of sample means will be approximately normal, regardless of the shape of the population distribution.

To calculate the probability, we need to convert the sample mean score to a z-score and then find the corresponding area under the standard normal distribution curve.

Calculate the standard error of the mean (SE):

SE = σ / sqrt(n)

SE = 127 / sqrt(72)

SE ≈ 14.96

Calculate the z-score:

z = (X - μ) / SE

z = (567 - 572) / 14.96

z ≈ -0.334

Find the probability corresponding to the z-score:

We can find the area to the left of the z-score -0.334. Let's denote this probability as P(z < -0.334).

The probability that the sample mean score is less than 567 is approximately equal to P(z < -0.334).

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Answer:

200%.

100% of 12 is 12 since for example if a coat is 10$ and 100% off then its now 0$. Therefore 200% is twice of 12 making it 24.

24=200% of 12

Step-by-step explanation:

There were total 276 guests in the wedding that Joy organised.

Given,

fraction of the guest that chose beef = 1/3

fraction of the guest that chose chicken = 5/12

fraction of the guest that chose vegetarian = 1  - 1/3 - 5/12 = 1/4

we are asked to find the total number of guests in the wedding:

guests that chose vegetarian = 1/4

guests that chose vegetarian = 69

1/4 of the guest = 69

4/4 of the guest = 69 x 4 = 276

Hence there are total 276 guests in the wedding.

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This statement is False.

Now let us see why:

To check whether the statement E[X^2]=4Var[X] is True or False for the given information, we need to recall the formulas of the expected value and variance of a Poisson distribution.

Equation of a Poisson distribution

P(X = k) = e^(-λ)*λ^(k)/k!, where k is the number of events in the given time interval, λ is the rate at which the events occur

Expected Value of a Poisson distribution:

E(X) = λ

Variance of a Poisson distribution:

Var(X) = λ

So, for a Poisson distribution, E(X^2) can be calculated as follows:

E(X^2) = λ + λ^2

Where, λ = average rate/ mean rate = 1/10 = 0.1

So, E(X^2) = 0.1 + 0.01 = 0.11

And Var(X) = λ = 0.1

Now, let's check whether the statement E[X^2]=4Var[X] is True or False

E[X^2] = 0.11 ≠ 4 * Var[X] = 0.4 (False)

Hence, the statement E[X^2]=4Var[X] is False.

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Answer:

4

Step-by-step explanation:

24/6=4

The equation that represents the total cost as a function of the number (in hundreds) of posters ordered, x is f(y) = 109.95 + 65x

let

y = total cost

x = numbers in hundreds

cost of first hundred = $109.95

Cost of additional hundred = $65

f(y) = 109.95 + 65x

The total cost of 1000 posters

1000 - first hundred = 900

additional hundreds = 900/100

= 9

So,

f(y) = 109.95 + 65x

= 109.95 + 65(9)

= 109.95 + 585

= 694.95

f(y) = $694.95

Therefore, the total cost of 1000 posters is f(y) = $694.95

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Answer:

Step-by-step explanation:

If the cost of 1 day lilies is 9 and the cost of bunch is 4

So ming spent = 3 x 9 + 4 x 8 = 59

Lisa spent = 5 x 9 + 4 x 8 = 77

So the cost of 1 day lilies is 9$ and 1 bunch is 4$

Answer:

X=23

Step-by-step explanation:

12²+5²=(X-10)²

144+25=(X-10)²

169=(X-10)²

√169=√(X-10)²

13=X-10

X=23

The value of the given function  f(x) = 25000 + 4000x, where x -4, 1.5 , 3.

=  9,000

=  31000

=  37000

A function is defined as a relationship between a group of inputs that each have one output. A function seems to be a relationship between inputs in which each input is associated with exactly one output. Every function seems to have a domain and a codomain, as well as a range.

The financial aid can be determined by the function f(x) = 25000 + 4000x.

When x = -4

f(x) = 25000 + 4000x

f(-4) = 25000 + 4000(-4)

        = 25000 - 16000.

        = 9,000

When x = 1.5

f(x) = 25000 + 4000x

f(1.5) = 25000 + 4000(1.5)

         = 25000 + 6000

        = 31000

When x = 3

f(x) = 25000 + 4000x

f(3) = 25000 + 4000(3)

      = 25000 + 12000

      = 37000

The value of the given function  f(x) = 25000 + 4000x, where x -4, 1.5 , 3.

=  9,000

=  31000

=  37000

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Answer:

4536

Step-by-step explanation:

yes

The new digital camera that creates photos that are best printed at 4 inches by 6 inches will minimize.

The ratio is defined as a relationship between two quantities, it is expressed as one divided by the other.

Since the aspect ratio of 4:6 is closer to that of an 11-inch by 14-inch frame (which is approximately 1:1.5) than the aspect ratio of 3.5:5 (which is approximately 7:10).

By minimizing the difference in aspect ratio between the photo and the frame, you will minimize the loss of the edges of the picture when trying to fit it into the frame.

The new digital camera that creates photos that are best printed at 4 inches by 6 inches will minimize the loss of the edges of the picture when trying to fit it into an 11-inch by 14-inch frame.

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The Linnaean class system provides a framework for understanding the diversity of life on Earth by grouping similar organisms together based on shared characteristics.

By comparing and contrasting the definitions of each class, we can see how different groups of animals are related to each other and how they differ in terms of their biological traits.

The Linnaean class system is a way of organizing living things based on shared characteristics. Let's compare and contrast the definitions of the Linnaean classes using a Venn diagram.

First, we have the class Mammalia, which includes all animals that have hair or fur, produce milk to feed their young, and have specialized teeth. This class overlaps with the class Aves, which includes all birds, because some birds have specialized beaks and feathers that are similar to mammalian hair and teeth. However, birds do not produce milk.

Next, we have the class Reptilia, which includes animals that are cold-blooded, lay eggs, and have scales or plates on their skin. This class overlaps with both Mammalia and Aves in terms of species that lay eggs, such as monotremes (platypus and echidnas) and some birds (ostriches and emus). However, reptiles lack specialized teeth and do not produce milk.

Finally, we have the class Amphibia, which includes animals that are cold-blooded, breathe through their skin, and undergo metamorphosis from a water-dwelling larval stage to a land-dwelling adult stage. This class overlaps with Reptilia in terms of some shared characteristics, but Amphibia also lacks specialized teeth and does not lay eggs with hard shells like reptiles.

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Based on the exponential decay equation, the number of employees for the company that has been decreasing yearly by 5%, will in 10 years be approximately 515.

The exponential decay equation or function gives the value in t years that has a constant ratio of decrease.

Exponential decay equation is one of the two exponential functions.  The other is the exponential growth equation.

The annual decrease in the number of employees = 5%

The current number of employees in the company = 860

The expected time = 10 years.

The exponential decay equation is as follows, y = 860 x 0.95^10.

y = 860 x 0.95^10 = 515

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The probability that a student buys coffee OR a muffin before class is 55%

The given parameters are:

P(Coffee) = 30%

P(Muffin) = 40%

P(Tea) = 20%

P(Coffee and Muffin) = 15%

P(Tea and Muffin) = 10%

The probability that a student buys coffee OR a muffin before class is

P(Coffee or Muffin) = P(Coffee) + P(Muffin) - P(Coffee and Muffin)

This gives

P(Coffee or Muffin) = 30% + 40% - 15%

Evaluate

P(Coffee or Muffin) = 55%

Hence, the probability that a student buys coffee OR a muffin before class is 55%

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Answer:

45

Step-by-step explanation:

Answer:

12

Step-by-step explanation:

Here the mixture (or solution) contains alcohol and water.

Given that you have 85% alcohol mixture. It means that each 1 mL of the mixture will contain 0.85 mL of alcohol, and the remaining 0.15 mL of pure water.

So the amount of alcohol in 120 mL of mixture is calculated as,

When 'x' mL of pure water is added to the solution, the total amount of water in the mixture becomes,

Since no alcohol is added, the amount of alcohol in the final mixture will remain same, that is, 102 mL.

Now, it is given that the concentration of the final mixture is 10%, So the amount of alcohol in the solution should be 10% of the total mixture,

Thus, you need to add 900 mL of pure water in order to have 10% alcohol solution.

And the equation used to solve this problem is,

The graph of f(x) = 3* has a y-intercept at (0, 3), an x-intercept at (1, 0), and a horizontal asymptote at y = 4. The exponential function f(x) = a* is increasing if a > 1 and decreasing if 0 < a < 1.

The y-intercept of the graph occurs when x = 0, so substituting x = 0 into the function f(x) = 3*, we get f(0) = 3*0 = 3. Therefore, the y-intercept is (0, 3).

The x-intercept of the graph occurs when y = 0, so substituting y = 0 into the function f(x) = 3*, we get 0 = 3*x. Solving for x, we find x = 1. Therefore, the x-intercept is (1, 0).

The horizontal asymptote represents the value that the function approaches as x approaches positive or negative infinity. In this case, the horizontal asymptote is y = 4, indicating that as x becomes extremely large or extremely small, the function approaches a value of 4.

For the exponential function f(x) = a*, the value of a determines whether the function is increasing or decreasing. If a > 1, the function is increasing as x increases. If 0 < a < 1, the function is decreasing as x increases.

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Answer:

i think it is -1

Step-by-step explanation:

Answer:

108 degrees

Step-by-step explanation:

hope that works

Answer:

a and c because the slopes sre reciprocals

Add/Subtract like terms

Answer:

2x^2-6x-5

Hope this helped :)

I typed the wrong answer and I cant delete so dont do it at all

The given message signal g(t) consists of multiple sinc and cosine components. It is sampled at a rate 25% higher than the Nyquist rate and quantized into L levels. The maximum acceptable error in sample amplitudes is limited to 0.1% of the peak signal amplitude.

To sketch the amplitude spectrum of g(t), we observe that sinc functions centered at 16 kHz and 10 kHz contribute amplitudes of 16x10³ and 10x10³, respectively, while the cosine component centered at 30 kHz has an amplitude of 20x10³. The horizontal axis represents the frequency (f).

The amplitude spectrum of the sampled signal, within the range -50 kHz to 30 kHz, will exhibit replicas of the original spectrum centered at multiples of the sampling frequency. The amplitudes and frequencies should be labeled according to the replicated components.

The minimum required bandwidth for binary transmission can be determined by considering the highest frequency component in g(t), which is 30 kHz. Therefore, the minimum required bandwidth will be 30 kHz.

For M-ary multi-amplitude signaling within a channel bandwidth of 50 kHz, we need to find the minimum value of M. It can be determined by comparing the available bandwidth with the required bandwidth for each amplitude component of g(t). The minimum M will be the smallest number of levels needed to represent all the significant amplitude components without violating the bandwidth constraint.

To minimize M, we need to select a pulse shape that achieves the narrowest bandwidth while maintaining an acceptable level of distortion. Different pulse shapes can be considered, such as rectangular, triangular, or raised cosine pulses.

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In the standard curve for nitrophenol generated for use in this experiment, the nitrophenol concentration is reported on the x - axis.

What is standard curve?

A calibration curve, sometimes referred to as a standard curve, is a common technique used in analytical chemistry to compare an unknown sample to a group of standard samples with known chemical concentrations.

In order to calculate the nitrophenol concentration in mole/ml based on absorbance readings, the standard curve is drawn.

Based on the optical density, or OD410, or absorbance at 410 nm, this standard curve of absorbance is used to calculate the quantity of nitrophenol.

Therefore, the nitrophenol concentration is displayed on the x axis of the standard curve.

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the derivative dy/dx is equal to 1/3 based on the given information. It's important to note that this calculation assumes that the partial derivatives (∂F/∂x) and (∂F/∂y) are not zero at the given point (z.y.a).

n the given problem, we have an implicit equation ln(x+y+z) = x+2y+3z that defines z as a function of x and y. We are given the values dy = (-1, 5, -3).

To find the derivative dy/dx, we can use the total derivative formula and apply it to the implicit equation. The total derivative is given by dy/dx = - (∂F/∂x)/(∂F/∂y), where F = ln(x+y+z) - x - 2y - 3z.

Differentiating F partially with respect to x and y, we have (∂F/∂x) = 1/(x+y+z) - 1 and (∂F/∂y) = 1/(x+y+z) - 2.

Plugging in the given values of dy = (-1, 5, -3), we can calculate dy/dx = - (∂F/∂x)/(∂F/∂y) = -(-1)/(5-2) = 1/3.

Therefore, the derivative dy/dx is equal to 1/3 based on the given information. It's important to note that this calculation assumes that the partial derivatives (∂F/∂x) and (∂F/∂y) are not zero at the given point (z.y.a).

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The quadratic approximation of the function f(x, y) = e^x ln(1 + y) near the origin is f_quadratic(x, y) = y, and the cubic approximation is f_cubic(x, y) = y.

To find the quadratic and cubic approximations of the function f(x, y) = e^x ln(1 + y) near the origin using Taylor's formula, we need to compute the partial derivatives of f with respect to x and y at the origin (0, 0) and evaluate the function and its derivatives at the origin.

First, let's compute the partial derivatives:

f_x(x, y) = (d/dx) (e^x ln(1 + y)) = e^x ln(1 + y)

f_y(x, y) = (d/dy) (e^x ln(1 + y)) = e^x / (1 + y)

Next, we evaluate the function and its derivatives at the origin:

f(0, 0) = e^0 ln(1 + 0) = 0

f_x(0, 0) = e^0 ln(1 + 0) = 0

f_y(0, 0) = e^0 / (1 + 0) = 1

Using these values, we can write the quadratic approximation of f near the origin as:

f_quadratic(x, y) = f(0, 0) + f_x(0, 0) * x + f_y(0, 0) * y = 0 + 0 * x + 1 * y = y

Similarly, we can find the cubic approximation:

f_cubic(x, y) = f(0, 0) + f_x(0, 0) * x + f_y(0, 0) * y + (1/2) * f_xx(0, 0) * x^2 + f_xy(0, 0) * x * y + (1/2) * f_yy(0, 0) * y^2

             = 0 + 0 * x + 1 * y + (1/2) * 0 * x^2 + 0 * x * y + (1/2) * 0 * y^2 = y

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Answer:

79-22=8h

Step-by-step explanation:

hope this helps!