After two numbers are removed from the list $$9,~13,~15,~17,~19,~23,~31,~49,$$ the average and the median each increase by $2$. What is the product of the two numbers that were removed?

Answer:

The very first one is decay and the rest are growth

Step-by-step explanation:

Im SO sorry if i got it wrong

I REALLY hope this helped

Best of luck

The coefficient of determination (R²) for the fitted least squares line is 0.75.

To fit the least squares line through the given points and find the coefficient of determination (R²), we can follow these steps:

Let's perform these calculations:

Step 1: Calculate the mean values of x and y.

x' = (1 + 2 + 3) / 3 = 2

y' = (1 + 1 + 3) / 3 = 5/3 ≈ 1.6667

Step 2: Calculate the sums of squares: SSxx, SSyy, and SSxy.

SSxx = Σ((xi - x')²) = (1 - 2)² + (2 - 2)² + (3 - 2)² = 2

SSyy = Σ((yi - y')²) = (1 - 5/3)² + (1 - 5/3)² + (3 - 5/3)² = 8/3 ≈ 2.6667

SSxy = Σ((xi - x')(yi - y')) = (1 - 2)(1 - 5/3) + (2 - 2)(1 - 5/3) + (3 - 2)(3 - 5/3) = 4/3 ≈ 1.3333

Step 3: Calculate the slope (m) and y-intercept (b) of the least squares line.

m = SSxy / SSxx = 1.3333 / 2 = 2/3 ≈ 0.6667

b = y' - mx' = 5/3 - (2/3)(2) = 5/3 - 4/3 = 1/3 ≈ 0.3333

Therefore, the equation of the least squares line is y = 0.6667x + 0.3333.

Step 4: Calculate the predicted y-values (y_pred) using the least squares line equation.

For (1, 1):

y_pred = 0.6667 × 1 + 0.3333 = 0.6667 + 0.3333 = 1

For (2, 1):

y_pred = 0.6667 × 2 + 0.3333 = 1.3334 + 0.3333 ≈ 1.6667

For (3, 3):

y_pred = 0.6667 × 3 + 0.3333 = 2 + 0.3333 ≈ 2.3333

The predicted y-values are (1, 1), (2, 1.6667), and (3, 2.3333).

Step 5: Calculate the residual sum of squares (RSS) and the total sum of squares (TSS).

RSS = Σ((yi - y_pred)²) = (1 - 1)² + (1 - 1.6667)² + (3 - 2.3333)² ≈ 0.6667

TSS = SSyy = 8/3 ≈ 2.6667

Step 6: Calculate the coefficient of determination (R²) using the formula: R² = 1 - (RSS / TSS).

R² = 1 - (0.6667 / 2.6667) = 1 - 0.25 = 0.75

Therefore, the coefficient of determination (R²) for the fitted least squares line is 0.75.

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Answer:

I will suppose that the actual question is:

Iodine-131 has a half-life of 8 days. How much would be left of an original g sample after x days?

Ok, a half-life means that after that time, the mass of the original sample is reduced to half.

So if we start it a quantity g of iodine-131, after 8 days, we will have g/2.

Also, remember that the decay is written as an exponential decay, then we will have:

A(x) = g*(r)^x

where:

A is the amount of the sample after x days, g is the initial amount of the material (such that A(0) = g) and r is the rate of decay.

We know that:

A(8) = g/2 = g*(r)^8

Now we can solve this for r:

g/2 = g*(r)^8

1/2 = r^8

(1/2)^(1/8) = r = 0.917

Then the amount of material after x days is given by:

A(x) = g*(0.917)^x

Answer: A

because It has 1 one 2 twos 3 threes 2 fours 1 five And 1 six

Answer:

28 and 12t

Step-by-step explanation:

4 x 7

4 x 3t

Answer:

28+12t

Step-by-step explanation:

Simplify the expression :)

btw you spelled please wrong

Answer: 95

Step-by-step explanation:

Think of 1/8 times 760 as 760/8 because it’s the same thing.

The renewal-type equation for E[SN(1)+1] is E[SN(1)+1] = 2, indicating that the expected value of the sum of the number of renewals by time 1 plus 1 is equal to 2.

To derive a renewal-type equation for E[SN(1)+1], we can use the renewal-reward theorem.

Let Tn be the interarrival times of the renewal process, where n represents the nth renewal. The random variable N(t) represents the number of renewals that occur by time t.

Using the renewal-reward theorem, we have:

E[SN(1)+1] = E[T1 + T2 + ... + TN(1) + 1]

Since the interarrival times are independent and identically distributed (i.i.d.), we can express this as:

E[SN(1)+1] = E[T] * E[N(1)] + 1

Now, we need to compute the expressions for E[T] and E[N(1)].

E[T] represents the expected interarrival time, which is equal to the reciprocal of the renewal rate. Let λ be the renewal rate, then E[T] = 1/λ.

E[N(1)] represents the expected number of renewals by time 1. This can be calculated using the renewal equation:

E[N(t)] = λ * t

Therefore, E[N(1)] = λ * 1 = λ.

Substituting these expressions back into the renewal-type equation, we have:

E[SN(1)+1] = (1/λ) * λ + 1 = 1 + 1 = 2

Hence, the renewal-type equation for E[SN(1)+1] is E[SN(1)+1] = 2.

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Answer:

Step-by-step explanation:

Answer:

77.28

Step-by-step explanation:

c=π2r

12.3 times 2 =

24.6π

=77.28317928

=77.28

Answer:

77.3

Step-by-step explanation:

The probability that the mean completion time of 4 projects falls between 16 and 19 months is approximately 0.6568 or 65.68%.

The Central Limit Theorem states that for a sufficiently large sample size, the distribution of sample means will approach a normal distribution regardless of the shape of the population distribution.

Specifically, if we have a random sample of n observations drawn from a population with mean μ and standard deviation σ, then the distribution of the sample means will have a mean equal to the population mean μ and a standard deviation equal to the population standard deviation σ divided by the square root of the sample size n.

In this case, the average time taken to complete a project in the real estate company is 18 months, with a standard deviation of 3 months.

Assuming that the project completion time approximately follows a normal distribution, we can use the Central Limit Theorem to find the probability that the mean completion time of 4 such projects falls between 16 and 19 months.

First, we need to calculate the standard deviation of the sample mean. Since we have 4 projects, the sample size is n = 4.

Therefore, the standard deviation of the sample mean is σ/√n = 3/√4 = 3/2 = 1.5 months.

Next, we can standardize the values of 16 and 19 months using the formula z = (x - μ) / (σ/√n), where x is the value, μ is the population mean, σ is the population standard deviation, and n is the sample size.

For 16 months: z1 = (16 - 18) / (1.5) = -2/1.5 = -1.33

For 19 months: z2 = (19 - 18) / (1.5) = 1/1.5 = 0.67

Using a standard normal distribution table, we can look up the probabilities corresponding to the z-scores -1.33 and 0.67.

The table provides the cumulative probabilities for values up to a certain z-score.

For -1.33, the cumulative probability is approximately 0.0918.

For 0.67, the cumulative probability is approximately 0.7486.

To find the probability between these two z-scores, we subtract the cumulative probability associated with -1.33 from the cumulative probability associated with 0.67:

P(-1.33 < Z < 0.67) = 0.7486 - 0.0918 = 0.6568

Therefore, the probability that the mean completion time of 4 projects falls between 16 and 19 months is approximately 0.6568 or 65.68%.

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Answer:

[tex]\theta=30.285^{\circ}[/tex]

Step-by-step explanation:

The range of a projectile is given by :

[tex]R=\dfrac{u^2\sin2\theta}{g}[/tex]

Put R = 20 m, u = 15 m/s and finding the value of angle of projection

So,

[tex]R=\dfrac{u^2\sin2\theta}{g}\\\\\sin2\theta=\dfrac{Rg}{u^2}\\\\\sin2\theta=\dfrac{20\times 9.8}{15^2}\\\\\sin2\theta=0.871\\\\2\theta=\sin^{-1}(0.871)\\\\2\theta=60.57\\\\\theta=30.285^{\circ}[/tex]

So, the required angle of projection is equal to [tex]30.285^{\circ}[/tex].

Answer:

13/6

Step-by-step explanation:

slope = (y2-y1)/(x2-x1) where the variables indicate the coordinates of the two points

slope = (-7-6)/(-5-1) = -13/-6 = 13/6

Answer:

A. The median for town A, 30 degrees, is less than the median for town B, 40 degrees.

Step-by-step explanation:

The probability of detecting at most one defective component when all components of the three panels are examined is approximately 0.78136 or 78.14%.

To calculate the probability of detecting at most one defective component when all components of the three panels are examined, we need to consider the possible combinations of defective components in each panel.

Let's break down the problem step by step:

Panel 1:

- There are 5 components in Panel 1.

- The probability of a component being defective is 0.09.

- We want to calculate the probability of detecting at most one defective component.

The probability of detecting no defective components in Panel 1 is:

P(0 defective) = (1 - 0.09)^5 = 0.52201

The probability of detecting exactly one defective component in Panel 1 is:

P(1 defective) = 5 * 0.09 * (1 - 0.09)^4 = 0.40408

The probability of detecting at most one defective component in Panel 1 is:

P(at most 1 defective) = P(0 defective) + P(1 defective) = 0.52201 + 0.40408 = 0.92609

Panel 2 and Panel 3 have the same probabilities as Panel 1 since they also have 5 components and the same probability of a component being defective.

Now, to calculate the probability of detecting at most one defective component when examining all three panels, we multiply the probabilities of each panel:

P(at most 1 defective in all three panels) = P(at most 1 defective in Panel 1) * P(at most 1 defective in Panel 2) * P(at most 1 defective in Panel 3)

                                          = 0.92609 * 0.92609 * 0.92609

                                          = 0.78136

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Answer:

The 1st and 4th statements are true.

Option (b) 2.8600 is the correct answer.

To find the number of rental trucks that a local U-Move moving truck rental company can expect to rent during a given week, we need to find the expected value of the probability distribution.

The expected value of a probability distribution is given by:

Expected Value = Sum of (Number of Rented Trucks × Probability)

Therefore, Expected Value = (0 × 0.23) + (1 × 0.18) + (4 × 0.27) + (5 × 0.32)

Expected Value = 0 + 0.18 + 1.08 + 1.6Expected Value = 2.86

Therefore, the company can expect to rent 2.86 rental trucks during a given week. Option (b) 2.8600 is the correct answer.

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Answer: 504. Multiple for : 18, 24 and 42. Factorize of the above numbers : 18 = 2 • 32 24 = 23 • 3. 42 = 2 • 3 • 7

For the given first-order differential equation, we need to determine the largest interval on which a unique solution exists for each initial condition. The interval will depend on the specific initial condition and the behavior of the differential equation.

The first-order differential equation is given as:

t^et y' + y' + t^2 – 25yt - 99

To determine the largest interval on which a unique solution exists for each initial condition, we need to consider the behavior of the equation and any possible singularities or discontinuities.

For each initial condition, we can use standard techniques such as separation of variables or integrating factors to solve the differential equation and find the solution. The solution will depend on the initial condition and may have different behaviors based on the values of t and y.

It's important to note that the existence and uniqueness of solutions are generally guaranteed within a certain interval as long as the equation and initial condition satisfy certain conditions, such as Lipschitz continuity. However, without specific initial conditions, it is not possible to determine the exact intervals on which a unique solution exists.

Therefore, to determine the largest interval on which a unique solution exists for each initial condition, further analysis and specific initial conditions are required to assess the behavior of the equation and identify any constraints or limitations on the solution.

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Answer:

a) 25/2 or 12.5

b) 78,125

c) 625

d) 30,517,578,125

Answer:

y=2(x−4)²+3

Step-by-step explanation:

Option D. The numbers are 20 and 0.

Let the two nonnegative numbers be x and y such that x + y = 20. We know that the sum of the squares of the two nonnegative numbers x and y is as large as possible and as small as possible.

x + y = 20, or y = 20 - x (Since the numbers are non-negative, x, y ≥ 0)

Substituting y = 20 - x into x² + y² = P (for the sake of simplicity), we get x² + (20 - x)² = Px² + 400 - 40x + x² = P

We will take the first derivative with respect to x now: 2x - 40 = 0x = 20

Therefore, one of the nonnegative numbers is 20, and the other is zero. Consequently, the smallest possible sum of squares is 400 (since 20² + 0² = 400).Option D. The numbers are 20 and 0.

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Given, the sum of two nonnegative numbers is 20.

The problem asks us to find the numbers if the sum of their squares is as large as possible; as small as possible.

Therefore, let's find the sum of their squares at first.If 'x' and 'y' are two numbers, then the sum of their squares is given by:

[tex]x^2 + y^2[/tex]

If the sum of two nonnegative numbers is 20, then one number can be written as x and the other number can be written as y in terms of x.

Thus,y = 20 − xNow, the sum of their squares:

[tex]x^2 + y^2 = x^2 + (20 - x)^2[/tex]
= [tex]x^2 + 400 + x^2 - 40x[/tex]
= [tex]2x^2 - 40x + 400[/tex]
The above expression represents a parabola which opens upward because the coefficient of x^2 is positive.

Therefore, the sum of the squares of the two numbers will be maximum at the vertex of the parabola.

The x-coordinate of the vertex can be found as

:−b/2a = −(−40)/(2.2) = 10Hence, x = 10 and y = 10.

Substituting x = 10 and y = 10, we get

[tex]x^2 + y^2 = 200.[/tex]

Now, to find the smallest value of the sum of their squares, we can observe that the smallest value of x is 0, and the largest value of y is 20.

Thus, if x = 0 and y = 20, we get x^2 + y^2 = 400.

Answer:  The numbers are 10 and 10. The numbers are 0 and 20.

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Answer:

36×18×18×27

Step-by-step explanation:

Assuming the picture is Figure A you would multiply value from figure A by 3 to get corresponding value for dilated figure.

So if figure A is

12 × 6 × 6 × 9

the dilated figure would be

36 × 18 × 18 × 27

To predict a linear regression score, you first need to train a linear regression model using a set of training data.

Once the model is trained, you can use it to make predictions on new data points. The predicted score will be based on the linear relationship between the input variables and the target variable,

A higher regression score indicates a better fit, while a lower score indicates a poorer fit.

To predict a linear regression score, follow these steps:

1. Gather your data: Collect the data p

points (x, y) for the variable you want to predict (y) based on the input variable (x).

2. Calculate the means: Find the mean of the x values (x) and the mean of the y values (y).

3. Calculate the slope (b1): Use the formula b1 = Σ[(xi - x)(yi - y)]  Σ(xi - x)^2, where xi and yi are the individual data points, and x and y are the means of x and y, respectively.

4. Calculate the intercept (b0): Use the formula b0 = y - b1 * x, where y is the mean of the y values and x is the mean of the x values.

5. Form the linear equation: The linear equation will be in the form y = b0 + b1 * x, where y is the predicted value, x is the input variable, and b0 and b1 are the intercept and slope, respectively.

6. Predict the linear regression score: Use the linear equation to predict the value of y for any given value of x by plugging in the x value into the equation. The resulting y value is your predicted linear regression score.

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Answer:

( 6, -1 )

Step-by-step explanation:

When you rotate 1 from the x axis by 90° it becomes -1 from the y axis.

When you rotate 6 by 9° from thr y axis, it becomes again 6 on the x axis

Your new x value is 6 and y is -1

So (6,-1)

Answer:

(-6, 1)

Step-by-step explanation:

To find the point obtained by rotating point P = (1, 6) counterclockwise by an angle of 90 degrees (r₉₀°), we can use the rotation formula:

x' = x * cos(θ) - y * sin(θ)

y' = x * sin(θ) + y * cos(θ)

In this case, θ is 90 degrees.

Substituting the values into the formula:

x' = 1 * cos(90°) - 6 * sin(90°)

y' = 1 * sin(90°) + 6 * cos(90°)

cos(90°) = 0 and sin(90°) = 1, so we have:

x' = 1 * 0 - 6 * 1 = -6

y' = 1 * 1 + 6 * 0 = 1

Therefore, r₉₀°(P) = (-6, 1). The point P = (1, 6) rotates counterclockwise by 90 degrees to the point (-6, 1).

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