The temperature during a very cold day is recorded every 2 hours for 12 hours. The data are given in the table below.

Time (hours) 6 8 10 12 14 16 18
Temperature (℃) 3.88 6.48 9.37 10.42 8.79 4.96 0.69
Which polynomial models these data?


C(x) = 0.167x^3 + 2.76x^2 - 16.91x + 38.87

C(x) = 0.0034x^4 - 0.167x^3 + 2.76x^2 - 16.91x + 38.87

C(x) = 0.167x^3 + 2.76x^2 - 16.91x

C(x) = 0.0034x^4 - 0.167x^3 + 2.76x^2 - 16.91x

The variance of X is 3/80.

Given the probability density function of X,

f(x) = {3x² for 0 < x < 1

{0 otherwise

We can use this to answer the following:

(a) Find P(X < 0.5)

To find P(X < 0.5), we need to integrate the density function from 0 to 0.5:

P(X < 0.5) = ∫[0,0.5] f(x) dx

= ∫[0,0.5] 3x² dx

= [x³]₀.₃

= 0.125

(b) Find the cumulative distribution function of X, F(x)

The cumulative distribution function (CDF) of X is given by:

F(x) = P(X ≤ x) = ∫[0,x] f(t) dt

If x ≤ 0, then F(x) = 0. If 0 < x ≤ 1, then

F(x) = ∫[0,x] f(t) dt

= ∫[0,x] 3t² dt

= [t³]₀.ₓ

= x³

If x > 1, then F(x) = 1. So, the CDF of X is:

F(x) = {0 if x ≤ 0

{x³ if 0 < x ≤ 1

{1 if x > 1

(c) Find the expected value of X, E(X)

The expected value of X is given by:

E(X) = ∫[−∞,∞] x f(x) dx

Since the density function f(x) is zero outside the interval [0,1], we can restrict the integration to this interval:

E(X) = ∫[0,1] x f(x) dx

= ∫[0,1] 3x³ dx

= [3/4 x⁴]₀.₁

= 3/4 * 1⁴ - 0

= 3/4

Therefore, the expected value of X is 3/4.

(d) Find the variance of X, Var(X)

The variance of X is given by:

Var(X) = E(X²) - [E(X)]²

We have already found E(X) in part (c). To find E(X²), we integrate x² times the density function:

E(X²) = ∫[0,1] x² f(x) dx

= ∫[0,1] 3x⁴ dx

= [3/5 x⁵]₀.₁

= 3/5 * 1⁵ - 0

= 3/5

Substituting into the formula for variance:

Var(X) = E(X²) - [E(X)]²

= 3/5 - (3/4)²

= 3/5 - 9/16

= 3/80

Therefore, the variance of X is 3/80.

Complete question: Let X be a random variable defined by the density function

[tex]$$f(x)=\left\{\begin{array}{cl}3 x^2 & 0 \leq x \leq 1 \\0 & \text { otherwise }\end{array}\right.$$[/tex]

Find

(a)[tex]$E(X)$[/tex]

(b) [tex]$E(3 X-2)$[/tex]

(c) [tex]$E\left(X^2\right)$[/tex]

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Answer:

I'm not quite sure if you are joking or not, but here is your answer:
(a) The point estimate for the population mean price (per 100 pounds) is x = $6.88. The standard deviation is known to be σ = $1.92. The sample size is n = 42. For a 90% confidence interval, the critical value is 1.645 (obtained from a t-distribution table with 41 degrees of freedom). The margin of error is:

Margin of error = Critical value x Standard error

Standard error = Standard deviation / sqrt(sample size) = 1.92 / sqrt(42) = 0.2968

Margin of error = 1.645 x 0.2968 = 0.4882 ≈ 0.49

The lower limit of the confidence interval is:

Lower limit = Point estimate - Margin of error = 6.88 - 0.49 = $6.39

The upper limit of the confidence interval is:

Upper limit = Point estimate + Margin of error = 6.88 + 0.49 = $7.37

Therefore, the 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop is $6.39 to $7.37. The margin of error is $0.49.

(b) The maximal error of estimate is E = 0.25. The confidence level is 90%. The standard deviation is known to be σ = $1.92. We need to find the sample size (n) that satisfies the following formula:

Margin of error = Critical value x Standard error

Standard error = Standard deviation / sqrt(sample size)

For a 90% confidence interval, the critical value is 1.645. Substituting the known values and solving for n, we get:

0.25 = 1.645 x (1.92 / sqrt(n))

sqrt(n) = (1.645 x 1.92) / 0.25

n = [(1.645 x 1.92) / 0.25]^2

n ≈ 113

Therefore, a sample size of 113 farming regions is necessary for a 90% confidence level with maximal error of estimate E = 0.25 for the mean price per 100 pounds of watermelon.

(c) The farm brings 15 tons of watermelon to market, which is equivalent to 30,000 pounds (since 1 ton = 2,000 pounds). The point estimate for the population mean cash value of this crop is unknown. We can use the confidence interval obtained in part (a) to estimate this interval. Multiplying the lower and upper limits by 300 (since 300 pounds of watermelon correspond to $6.88), we get:

Lower limit = 6.39 x 300 = $1917

Upper limit = 7.37 x 300 = $2211

Therefore, the 90% confidence interval for the population mean cash value of this crop is $1917 to $2211. The margin of error is:

Margin of error = (Upper limit - Lower limit) / 2 = (2211 - 1917) / 2 = $147

The false statement is (a) a chi-square distribution with k degrees of freedom is more right-skewed than a chi-square distribution with k+1 degrees of freedom.

In fact, as the degrees of freedom increase, the chi-square distribution becomes less skewed and approaches a normal distribution. Statement (b) is true, a chi-square distribution never takes negative values. Statement (c) is generally true, the degrees of freedom for a chi-square test are determined by the sample size minus one. Statement (d) is incomplete, as there is no specified value for df. The larger the degrees of freedom, the smaller the p-value for a given chi-square value.

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I think it's 9, tell me if I'm wrong

a. If you have a good credit rating and can afford monthly payments of $586, you can borrow $32,521.48 from E-Loan. This can be calculated using the formula for a present value annuity:

PV = PMT x ((1 - (1 + r/n)^(-nt)) / (r/n))

Where PV is the present value, PMT is the monthly payment, r is the annual interest rate (3.9%), n is the number of times the interest is compounded per year (12 for monthly compounding), and t is the number of years (5 for a 60-month loan). Plugging in these values, we get:

PV = $586 x ((1 - (1 + 0.039/12)^(-12*5)) / (0.039/12)) = $32,521.48

b. The total interest you will pay for this loan is $3,911.88. This can be calculated using the formula for total interest paid on a loan:

Total interest = (PMT x n x t) - PV

Where PMT, n, and t are the same as before, and PV is the amount borrowed. Plugging in the values, we get:

Total interest = ($586 x 60 x 5) - $32,521.48 = $3,911.88

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The surface area of the sphere is 64π square units.

Option B is the correct answer.

We have,

Surface area of a sphere = 4πr² ______(1)

Now,

Radius = 4 units

Substituting in (1)

The surface area of a sphere

= 4πr²

= 4 x π x 4²

= 4π x 16

= 64π square units

Thus,

The surface area of the sphere is 64π square units.

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If in a random sample of n=125 units of the drug, there is an average dose of x=499.3 mg with a standard deviation of =6 mg the likelihood of the drugs produced containing a dosage of 500 mg is fairly high.

Based on the information provided, we can use the concept of the standard error of the mean to determine the likelihood that the drugs produced will contain a dosage of 500 mg.

The formula for the standard error of the mean is:

SE = s/√n

Where:
s = standard deviation of the sample
n = sample size

Substituting the values given, we get:

SE = 6/√125
SE = 0.54

This means that the sample mean of 499.3 mg is 0.54 units away from the true population mean of 500 mg.

To determine the likelihood of the drugs produced containing a dosage of 500 mg, we can use a confidence interval. A 95% confidence interval for the mean dosage can be calculated as:

Mean dosage ± 1.96(SE)

Substituting the values given, we get:

499.3 ± 1.96(0.54)
499.3 ± 1.06

The 95% confidence interval for the mean dosage is (498.24, 500.36).

Therefore, there is a 95% chance that the true population means dosage falls within this interval. Since the interval includes the value of 500 mg, we can conclude that the likelihood of the drugs produced containing a dosage of 500 mg is fairly high.

In a random sample of n=125 units of the drug, there is an average dose of x=499.3 mg with a standard deviation of =6 mg is very high.

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The solution to the given initial value problem is:

y1(t) = 3t - 2sin(2t) + 11/10sinh(t) - 11/10sinh(2t)

y2(t) = 6t + 7/10cosh(t) - 17/10sinh(t)

Taking the Laplace transform of the given system of differential equations, we get:

sY1(s) - y1(0) = 5Y1(s) - 4Y2(s) - 2(2/(s^3)) + 2(1/(s^2))

sY2(s) - y2(0) = 10Y1(s) - 77(1/s) - 17(1/(s^2)) - 2(1/(s^2))

Applying the initial conditions, we get:

sY1(s) - 3 = 5Y1(s) - 4Y2(s) - 4/s^3 + 2/s^2

sY2(s) = 10Y1(s) - 77/s - 17/s^2 - 2/s^2

Solving for Y2(s), we get:

Y2(s) = (10Y1(s) - 77/s - 17/s^2 - 2/s^2)/s

Substituting this in the equation for Y1(s), we get:

sY1(s) - 3 = 5Y1(s) - 4[(10Y1(s) - 77/s - 17/s^2 - 2/s^2)/s] - 4/s^3 + 2/s^2

Simplifying and solving for Y1(s), we get:

Y1(s) = (3s^3 + 10s^2 + 8s + 154)/(s^5 + 5s^3 + 4s)

Taking the inverse Laplace transform, we get:

y1(t) = 3t - 2sin(2t) + 11/10sinh(t) - 11/10sinh(2t)

y2(t) = 6t + 7/10cosh(t) - 17/10sinh(t)

Therefore, the solution to the given initial value problem is:

y1(t) = 3t - 2sin(2t) + 11/10sinh(t) - 11/10sinh(2t)

y2(t) = 6t + 7/10cosh(t) - 17/10sinh(t)

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The Quasilinearization Method is a technique used to approximate solutions for nonlinear differential equations by linearizing them iteratively. It is particularly helpful when solving first-order IVPs.

A maximal solution to a first-order IVP is a solution that exists on the largest possible interval, while a minimal solution exists on the smallest possible interval.

Example 1:
Consider the first-order IVP: dy/dt = y^2, y(0) = 1.

Maximal solution: The maximal solution to this IVP is y(t) = 1/(1 - t) on the interval (-∞, 1).

Minimal solution: The minimal solution is the same as the maximal solution for this example, as there are no other solutions that exist on a smaller interval.

Example 2:
Consider the first-order IVP: dy/dx = x + y, y(0) = 0.

Maximal solution: The maximal solution to this IVP is y(x) = -x + e^x - 1 on the interval (-∞, +∞).

Minimal solution: The minimal solution is the same as the maximal solution for this example since there are no other solutions that exist on a smaller interval.

In both examples, the Quasilinearization Method can be applied to linearize the differential equations and approximate the solutions. The maximal and minimal solutions represent the largest and smallest possible intervals where the solutions are valid.

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Educational psychology provides teachers with research-based principles to guide their teaching.

When teachers go through educational psychology, they are taught on ways to improve their teaching.

These ways will be based on research overtime that have proved efficient in helping students learn from teachers.

Some of these include empowering school social and cultural structures, minimizing bias, implementing an equity pedagogy, the method of knowledge creation, and integrating content (Banks, 1995a).

The main goal of multicultural education is to reduce barriers to educational opportunity and success for students from different cultural backgrounds. The principle that all pupils, regardless of culture, deserve educational equity serves as its cornerstone.

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The value of the probability P(3) is 1/4.

We have,

There are 4 outcomes.

i.e

3, 4, 5, and 6.

Now,

P(3)

This means,

The probability of getting 3 as the outcome when spun.

So,

P(3) = 1/4

Thus,

The value of the probability P(3) is 1/4.

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The amount of paper used for the label on the can of tune is 12.57 in²

Here, the shape of the can of can is cylindrical.

The area of the cured surface of cylinder is given by formula,

A = 2πrh

where r is the radius of the cylinder

and h is the height of the cylinder

Here, r = 2 in and h = 1 in

so, the area of the lateral surface of cylinder would be,

A = 2 × π × r × h

A = 2 × π × 2 × 1

A = 4 × π

A = 12.57 sq. in.

Therefore, the required amount of paper = 12.57 in²

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Answer:

BE = 7.73

Step-by-step explanation:

All the angles in ΔABC are 60 because it's an equilateral triangle

AB = AE = 4

m∠BAE = 60 + 90 = 150

(1/2)m∠BAE = 150/2  = 75

sin75 = (1/2)(BE) / 4

1/2(BE) = sin75(4)

BE = sin75(4)(2) = 7.73

Answer:

The correct answer is $6,373.02.

We can use the formula for present value of an annuity:

PV = PMT x ((1 - (1 + r/n)^(-n*t)) / (r/n))

Where PV is the present value, PMT is the monthly payment, r is the annual interest rate, n is the number of times the interest is compounded per year, and t is the number of years.

Plugging in the values, we get:

PV = 300 x ((1 - (1 + 0.12/12)^(-12*2)) / (0.12/12))

PV = $6,373.02

Therefore, the present value of Maria's investment is $6,373.02.

The function f(x) = 3x³ - 2x² - 2x + 3 has one root and the root is x = 1

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

f(x) = 3x³ - 2x² - 2x + 3

Next, we plot the graph of the function f(x)

See attachment

From the graph, we can see that the graph intersects with the x-axis once at x = -1

This means that it has one root and the root is x = 1

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If right is a scaled copy of the rectangle on the left then the scale factor is  1/2.

The scale factor can be calculated by dividing the corresponding lengths (or widths) of the two rectangles.

The length of the left rectangle is 20 units, and the length of the right rectangle is 10 units.

Therefore, the scale factor for the length is:

scale factor for length = length of right rectangle / length of left rectangle

= 10 / 20

= 0.5

scale factor for width = width of right rectangle / width of left rectangle

= 5 / 10

= 0.5

Since the two scale factors are the same, we can conclude that the rectangles are scaled by the same factor of 0.5 in both the length and the width.

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After 10 days, the colony would be as large as 989, based on the exponential growth of 67% every 2 days.

An exponential growth refers to a constant rate or percentage of growth in the number or value of some variables.

Exponential growth can be modeled using the exponential growth function and used to determine the future quantity or amount of the variables.

Initial number of the bacteria microorganisms = 55

Growth rate = 67% every 2 days

Daily growth rate = 33.5% (67% ÷ 2)

The number of days involved, t = 10 days

The ending number of the bacteria microorganisms = Future Amount = 55(1 + 33.5%)^t

= 55(1.335)^10

= 989

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Answer:

55 0.67 ^5

Step-by-step explanation:

The future amount = ? micro organisms

The solution to the system of equations is x=-3.35, y=-7, z=3 using Cramer's method.

| 1  -1  2 |   | x |   | 7 |

| 0  -1  6 | x | y | = |17 |

| 3  -2  2 |   | z |   |14 |

We can use Cramer's rule to solve this system of equations by finding the determinants of the coefficient matrix and the matrices obtained by replacing each column with the constant terms.

The determinant of the coefficient matrix is:

| 1  -1  2 |

| 0  -1  6 |

| 3  -2  2 |

= 1(-1*2 - 6*(-2)) - (-1*2 - 6*3) + 2*(2*(-1) - (-1)*(-2))

= 20

The determinant obtained by replacing the first column with the constant terms is:

| 7  -1  2 |

|17  -1  6 |

|14  -2  2 |

= 7(-1*2 - 6*(-2)) - (-1*17 - 6*14) + 2*(2*(-1) - (-1)*(-2))

= -67

The determinant obtained by replacing the second column with the constant terms is:

| 1  7  2 |

| 0 17  6 |

| 3 14  2 |

= 1(17*2 - 6*14) - 7(3*2 - 14*2) + 2(3*17 - 14*0)

= -140

The determinant obtained by replacing the third column with the constant terms is:

| 1  -1  7 |

| 0  -1 17 |

| 3  -2 14 |

= 1(-1*14 - 17*(-2)) - (-1*7 - 17*3) + 7*(2*(-2) - (-1)*(-2))

= 60

Therefore, the solution to the system of equations is:

makefile

Copy code

x = -67/20

y = -140/20

z = 60/20

x = -3.35

y = -7

z = 3

Hence, the solution to the system of equations is x=-3.35, y=-7, z=3 using Cramer's method.

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Calculating Delta Chi-Square, Delta, Deviance, and Delta Beta is done using C. Jackknifing, like we used in MLR (Multiple Linear Regression).

Jackknifing is a resampling technique that helps to estimate the stability and accuracy of statistical measures. In this method, one observation is removed at a time, and the model is recalculated to determine the impact of that observation on the overall result. This process is repeated for each observation, which helps to assess the influence of each data point on the model's performance.

Delta Chi-Square is a measure of the change in the goodness of fit of the model when a variable is added or removed. Delta measures the change in the estimated coefficient of a variable when another variable is added or removed from the model. Deviance measures the difference between the log-likelihood of the model and the log-likelihood of the saturated model, which is the model that perfectly fits the data.

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In this case, the mean is 5.5 cm and the standard deviation is 0.1 cm. So, one standard deviation below the mean is 5.4 cm and one standard deviation above the mean is 5.6 cm. Therefore, about 68% of the data falls between 5.4 cm and 5.6 cm, which includes the range of 5.3 cm to 5.7 cm.

Your question involves a normal distribution with a mean (µ) of 5.5 cm and a standard deviation (σ) of 0.1 cm. You want to find the percentage of data between 5.3 cm and 5.7 cm.

First, we need to standardize the scores using the z-score formula: z = (x - µ) / σ

For 5.3 cm: z1 = (5.3 - 5.5) / 0.1 = -2
For 5.7 cm: z2 = (5.7 - 5.5) / 0.1 = 2

Now, referring to a standard normal distribution table or using a calculator, we find the area between these z-scores:

This can be determined using the empirical rule, also known as the 68-95-99.7 rule, which states that for a normal distribution:
- About 68% of the data falls within one standard deviation of the mean
- About 95% of the data falls within two standard deviations of the mean
- About 99.7% of the data falls within three standard deviations of the mean

P(-2 < z < 2) = P(z < 2) - P(z < -2) = 0.9772 - 0.0228 = 0.9544

Converting it to a percentage, we get 95.44%, which is approximately 95%.

So, the answer is A. 95%.

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81 is the largest number of two-digit which is a perfect square.

Perfect squares are those integers that result from multiplying any number by itself.

1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 are the numbers.

Therefore, there exist 17 two-digit numbers whose digit sums are squares.

The greatest number among those that form a perfect square must be found.

We are aware that the initial number of perfect squares is 100. Furthermore, it is a perfect square of 10.

The number whose square will appear is obviously going to be fewer than 10 today.

The square of 9, which is written as 9²=81, is the first number before 10, so let's start there.

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We can be 96% confident that the true mean weight of all women lies between 129.21 and 367.11.

The best point estimate of the mean weight of all women can be calculated using the formula:

Point estimate = sample mean = (sum of sample weights) / sample size

Here, the sample size is not given, so we cannot calculate the sample mean directly. However, we are given two sample statistics: the sample starting point (2) and the sample statistic (s) which is the sample standard deviation.

We can use the formula for the t-distribution to estimate the population mean:

t = (x - μ) / (s / √n)

where x is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size.

To find the point estimate, we can rearrange this formula to solve for x:

x = μ + t(s / √n)

Since we don't know the population mean μ, we will use the sample starting point 2 as an estimate. We also know the sample standard deviation s = 30.766 and we are given a 96% confidence interval, so we need to find the critical value of t for a two-tailed test with 96% confidence and degrees of freedom (df) = n - 1.

Using a t-distribution table or calculator, we find that the critical value for df = n - 1 = 1 is t = 12.71.

Plugging in the values, we get:

2 + 12.71 * (30.766 / √n) = x

Solving for x, we get:

x = 2 + 12.71 * (30.766 / √n)

We still need to find the sample size n in order to calculate the point estimate. We can use the sample statistic given, which is the sample standard deviation s = 30.766, to estimate the sample size using the formula:

s = √[(n-1)/n] * σ

where σ is the population standard deviation.

Plugging in the values, we get:

30.766 = √[(n-1)/n] * 30.766

Solving for n, we get:

n = 2.24

This suggests that the sample size is quite small, which may limit the accuracy of our point estimate.

Plugging in the value of n, we get:

x = 2 + 12.71 * (30.766 / √2.24)

x = 2 + 12.71 * 19.398

x = 248.16

Therefore, the best point estimate of the mean weight of all women is 248.16.

b. To find a 96% confidence interval for the mean weight of all women, we can use the formula:

CI = x ± t(α/2, df) * (s / √n)

where x is the point estimate, t(α/2, df) is the critical value for a two-tailed test with α = 0.04 and df = n - 1, s is the sample standard deviation, and n is the sample size.

Plugging in the values, we get:

CI = 248.16 ± 12.71 * (30.766 / √2.24)

CI = 248.16 ± 118.95

CI = (129.21, 367.11)

Therefore, we can be 96% confident that the true mean weight of all women lies between 129.21 and 367.11.

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If a woman pays $2.78 for some bananas and eggs. If each banana costs $0.69 and each egg costs $0.35, then she bought 4 bananas and 6 eggs.

Let's assume the woman bought x bananas and y eggs.

According to the problem, each banana costs $0.69 and each egg costs $0.35.

So the cost of x bananas would be 0.69x and the cost of y eggs would be 0.35y.

The total cost of the bananas and eggs is given as $2.78. So we can write the equation:

0.69x + 0.35y = 2.78

Now we need to solve for x and y.

We can start by multiplying the entire equation by 100 to get rid of the decimals:

69x + 35y = 278

We can also simplify the equation by dividing both sides by 1 (which doesn't change the equation):

69x/1 + 35y/1 = 278/1

Now we can use a system of equations to solve for x and y.

Let's solve for y in terms of x by isolating y on one side of the equation:

35y = 278 - 69x

y = (278 - 69x)/35

Now we can substitute this expression for y into the original equation:

0.69x + 0.35((278 - 69x)/35) = 2.78

Simplifying this equation, we get:

0.69x + 8 - 2x = 2.78

Solving for x, we get:

0.69x - 2x = 2.78 - 8

-1.31x = -5.22

x = 4

So the woman bought 4 bananas.

Now we can substitute this value for x into the expression we derived for y:

y = (278 - 69(4))/35

y = 6

So the woman bought 6 eggs.

Therefore, the woman bought 4 bananas and 6 eggs.

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To add polynomials in a horizontal method, combine like terms. For the vertical method, write the polynomials in standard form and align like terms in columns, and combine like terms. To subtract polynomials in a horizontal method, find the negative (opposite)  of the polynomial that is being subtracted and then combine like terms. For the vertical method, write the polynomials in standard form, align like terms, and subtract by adding the negative (opposite).

To add polynomials vertically,  one need to write them in standard form and align like terms in the columns. Combine like terms and add them to the polynomial.

Therefore, note that Polynomials are seen as expressions with variables and coefficients. Combine or subtract like terms when adding or subtracting them.

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WRITE Describe how to add and subtract polynomials using both the vertical and horizontal methods.

To add polynomials in a horizontal method, combine  ----- For the vertical method, write the polynomials in  --------  align like terms in columns, and combine like terms. To subtract polynomials in a horizontal method, find the ------ of the polynomial that is being and then combine like terms. For the vertical method, write the polynomials in standard form, align like terms and  and subtract by adding the  -------

The given statement, "The mean income of workers who have majored in history is less than $25,000," is an alternative hypothesis.

An alternative hypothesis is a statement that is tested against the null hypothesis, which generally claims no relationship or effect. In this case, the null hypothesis would be, "The mean income of workers who have majored in history is greater than or equal to $25,000."

We must first comprehend the idea of null and alternative hypotheses in hypothesis testing before we can comprehend why the above statement is an alternate hypothesis.

When doing a hypothesis test, we begin with a null hypothesis, which is a claim that there is no connection between the variables under investigation.

The null hypothesis in this situation would be "The mean income of workers with history majors is greater than or equal to $25,000."

On the other side, the competing hypothesis asserts that the median income of people who majored in history is less than $25,000.

The supplied claim, "The mean income of workers who majored in history is less than $25,000," is, therefore, a counterclaim.

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Answer:I can't see the question

Step-by-step explanation:

a. The probability of winning $5 is when at least one dice comes up 6, which is 1 - 125/216 = 91/216.

b. The standard deviation of X is the square root of the variance:

SD(X) = √(9.828) = 3.135

c. The average profit over 100 rounds, X, will be approximately normally distributed with mean μ = E(X) = 0.6944 and standard deviation σ = SD(X)/√(n) = 3.135/√(100) = 0.3135.

d. The probability that Y is at most $80 is approximately 0.6325.

Probability is a measure of how likely an event is to occur. Many events are impossible to forecast with absolute accuracy. We can only anticipate the possibility of an event occurring, i.e. how probable they are to occur, using it.

(a) The probability distribution of X can be represented by the following table:

| X      | -1    | 5     |

|--------|------|-------|

| P(X=x) | 125/216  | 91/216 |

The probability of losing $1 is when none of the dice comes up 6, which is (5/6) x (5/6) x (5/6) = 125/216. The probability of winning $5 is when at least one dice comes up 6, which is 1 - 125/216 = 91/216.

(b) The expectation of X can be calculated as:

E(X) = (-1) x (125/216) + (5) x (91/216) = 0.6944

The variance of X can be calculated as:

Var(X) = [(−1 − 0.6944)² × 125/216] + [(5 − 0.6944)² × 91/216] = 9.828

The standard deviation of X is the square root of the variance:

SD(X) = √(9.828) = 3.135

(c) By the Central Limit Theorem, the average profit over 100 rounds, X, will be approximately normally distributed with mean μ = E(X) = 0.6944 and standard deviation σ = SD(X)/√(n) = 3.135/√(100) = 0.3135.

(d) Let Y be the total profit over 100 rounds. Then Y is the sum of 100 independent and identically distributed random variables with the same probability distribution as X. Therefore, by the Central Limit Theorem, Y is approximately normally distributed with mean μ_Y = 100μ = 69.44 and standard deviation σ_Y = √(100)σ = 31.35.

To find the probability that Y is at most $80, we standardize the variable:

Z = (80 - μ_Y)/σ_Y = (80 - 69.44)/31.35 = 0.337

Using a standard normal distribution table or calculator, we find that the probability of Z being less than or equal to 0.337 is 0.6325. Therefore, the probability that Y is at most $80 is approximately 0.6325.

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LCM stands for "Least Common Multiple". It is the smallest positive integer that is a multiple of two or more numbers

To find a formula for the least common multiple (LCM) of three positive integers a, b, and c.

The formula for LCM(a, b, c) is:
LCM(a, b, c) = LCM(LCM(a, b), c)

To find the LCM of two numbers, you can use the formula:
LCM(a, b) = (a * b) / GCD(a, b)
where GCD(a, b) is the greatest common divisor of a and b.

So, to find the LCM(a, b, c), follow these steps:

1. Calculate GCD(a, b)
2. Calculate LCM(a, b) = (a * b) / GCD(a, b)
3. Calculate GCD(LCM(a, b), c)
4. Calculate LCM(a, b, c) = LCM(LCM(a, b), c) = (LCM(a, b) * c) / GCD(LCM(a, b), c)

That's the formula and the step-by-step process to find the LCM of three positive integers a, b, and c.

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Answer:

The Answer Is B "Graph 2 has a larger sample standard deviation than Graph 1"

Step-by-step explanation:

Because the message in the text said the mean of Group 1's data is 2.57 while the sample mean of Group 2's data is 3.337 and from what I understand. 3.337 is a lot more than 2.57

Answer:

  (b)  Graph 2 has a larger sample standard deviation than Graph 1

Step-by-step explanation:

Given the two histograms of hours practiced, you want to know the relationship between the sample standard deviations of the two data sets.

The standard deviation is a measure of data variability. It will tend to be larger for less-symmetrical data distributions, and for those that are skewed one way or another.

The data of Graph 2 is less symmetrical than that of Graph 1, so we expect its standard deviation to be higher. A computation of the standard deviation confirms this.

Graph 1 standard deviation: about 1.40

Graph 2 standard deviation: about 1.53

Graph 2 has a larger sample standard deviation than Graph 1, choice B.

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Additional comment

In the computation, the first list (L1) is the set of data values. The second list, {2, 3, 4, ...} for example, is their relative frequencies—the heights of the bars in the histogram.

The given mean values seem to show that each bar is represented by its midpoint value, 0.5 for the first bar, for example. For the purpose of the standard deviation calculation, we don't need to make that adjustment.

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