Solve the non-linear ODE y"' +2/3 y' + only. y'=0 1 Y(1)=1 and y([infinity]) = 0

Answer:

Its the first one

Step-by-step explanation:

correct answer

The possible values of the coefficient (a) and an exponent (b) are CThe missing coefficient (a) can be any multiple of 3. The missing exponent (b) must be 5.

The two monomials are given as

ax³ 6xᵇ

Such that we have the LCM to be

LCM = 18x⁵

Since the coefficient of the LCM is 18, then the following is possible

a * 6 = multiples of 18

Divide both sides by 6

a = multiples of 3

Next, we have

LCM of x³ * xᵇ = x⁵

So, we have

b = 5 (bigger exponent)

Hence, the true statement is (c)

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a) The quadratic equation x² + 7x + 10 = 0 has two distinct real roots
b) The quadratic equation 4x² - 3x + 4 = 0 has two complex (non-real) roots.

The discriminant of a quadratic equation of the form ax² + bx + c = 0 is given by the expression b² - 4ac. The value of the discriminant can help us determine the nature of the roots of the quadratic equation.

Specifically:

If the discriminant is positive, then the quadratic equation has two distinct real roots.

If the discriminant is zero, then the quadratic equation has one real root (also known as a double root or a repeated root).

If the discriminant is negative, then the quadratic equation has two complex (non-real) roots.

Using this information, we can determine the number of real solutions for each of the given quadratic equations without actually solving them:\

a) x² + 7x + 10 = 0

Here, a = 1, b = 7, and c = 10.

Therefore, the discriminant is:

b² - 4ac = 7² - 4(1)(10) = 49 - 40 = 9

Since the discriminant is positive, this quadratic equation has two distinct real roots.

b) 4x² - 3x + 4 = 0

Here, a = 4, b = -3, and c = 4.

Therefore, the discriminant is:

b² - 4ac = (-3)² - 4(4)(4) = 9 - 64 = -55

Since the discriminant is negative, this quadratic equation has two complex (non-real) roots.

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The intervals over which  it is increasing or decreasing is:

Increasing on: ([tex]-\infty[/tex], -1)

Decreasing on: (-1, [tex]\infty[/tex])

The definitions for increasing and decreasing intervals are given below.

The function is :

[tex]f(x)=-x^2-2x+3[/tex]

We have to find the interval of function is decrease/increase .

Now, We have to first differentiate with respect to x , then:

f'(x) = - 2x + 2

This derivative is never 0 for real x.

In order to determine the intervals over which  it is increasing or decreasing.

Increasing on: ([tex]-\infty[/tex], -1)

Decreasing on: (-1, [tex]\infty[/tex])

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The graph of the function f(x)=|x−1|−3 is the graph (b)

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

f(x)=|x−1|−3

Express properly

So, we have

f(x) = |x − 1| − 3

The above expression is a absolute value function

This means that

Using the above as a guide, we have the following:

The graph of the function is the graph b

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Answer: x ≅ 0.5, and x ≅ −3.7

Step-by-step explanation:

[tex]4x^2 + 11x - 19 = -2x - 12[/tex]

[tex]4x^2 + 13x - 7 = 0[/tex]

Now use the quadratic formula with a = 4, b = 13, and c = -7

(if you dont know what that is, you should probably search it and understand/memorize).

Using the formula, we get two values:

x ≅ 0.5, and x ≅ −3.7

Step-by-step explanation:

Shift to the L 7 units    ( x+7)^2

Shift down 4       (x+7)^2    - 4  

For example, if the events were "you earn an A" and "your friend, who always studies with you, earns an A", these events would be dependent because the probability of your friend earning an A would be affected by whether or not you earn an A.

In this case, the events are not mutually exclusive because both events can happen at the same time (i.e., both you and a randomly selected student can earn an A in the course).

The events can be considered independent if one event does not affect the probability of the other event occurring. In this case, whether you earn an A does not affect the probability of the randomly selected student also earning an A. Therefore, the events can be considered independent.

Note that if the events were dependent, it would mean that the probability of one event occurring would affect the probability of the other event occurring. For example, if the events were "you earn an A" and "your friend, who always studies with you, earns an A", these events would be dependent because the probability of your friend earning an A would be affected by whether or not you earn an A.

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The equation of the plane passing through point P with normal vector 2i + 6j + 7k is x + 3y + 4z = 48.

A: True.

The equation of a plane in 3D space is given by Ax + By + Cz = D, where A, B, C are the components of the normal vector and D is the distance from the origin to the plane along the direction of the normal vector.

In this case, the normal vector is 2i + 6j + 7k, so A = 2, B = 6, and C = 7. To find D, we can substitute the coordinates of the given point P into the equation of the plane:

2(1) + 6(3) + 7(4) = D

2 + 18 + 28 = D

D = 48

Therefore, the equation of the plane passing through point P with normal vector 2i + 6j + 7k is x + 3y + 4z = 48.

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a) The probability that X is within three standard deviations of the mean is approximately 1.

b) the probability that the drone will land out of the targeted area exactly 4 times is 0.00052.

c) The probability that the drone will land out of the targeted area at most 4 times is 0.1029

d) The expected value of X is 9.6.

e) The meaning of the expected value in the context of the story is average landing performance of the drone based on the given probability of success.

f) The variance of X is 0.7319.

g) The probability that it missed the target at most 2 times is 3.121.

h) The probability that it missed the target at least 2 times is 0.7319.

I) The probability that X is within three standard deviations of the mean is 1.3856.

The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent trials with a constant probability of success.

The characteristics of a binomial random variable include the number of trials (n), the probability of success (p), the number of successes (x), and the mean and variance of the distribution.

Here we have

Binomial distribution We are testing the landing performance of a new automated drone. The drone lands on the targeted area 80% of the time. We test the drone 12 times.

a. X is a binomial random variable because we have a fixed number of independent trials and each landing has only two possible outcomes (landing on the targeted area or landing outside of it) with a constant probability of success (0.8).

The characteristics of the binomial distribution are:

The number of trials is fixed (n=12)

Each trial has only two possible outcomes (success or failure)

The probability of success (p) is constant for each trial

The trials are independent of each other

b. P(X = 4) = (12 choose 4) × (0.8)⁴ × (0.2)⁸ = 0.00052

c. P(X< = 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= 0.0687 + 0.2060 + 0.3020 + 0.2670 + 0.1854 = 0.1029

d. E(X) = np = 120.8 = 9.6

e. The expected value of X represents the average number of successful landings (in the targeted area) we would expect to see in a sample of 12 landings.

In the context of the story, it tells us the average landing performance of the drone based on the given probability of success.

f. Var(X) = np(1-p) = 120.80.2 = 1.92

g. P(X<=2 | X<=4) = P(X<=2)/P(X<=4)

= (P(X=0) + P(X=1) + P(X=2))/(P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4))

= 0.3217/0.1029 = 3.121

h. P(X>=2 | X<=4) = 1 - P(X<2 | X<=4) = 1 - P(X<=1 | X<=4) = 1 - (P(X=0) + P(X=1))/(P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)) = 1 - 0.2747/0.1029 = 0.7319

i. The standard deviation of a binomial distribution is √(np(1-p)). So, the standard deviation of X is √(120.80.2) = 1.3856. Three standard deviations above and below the mean would be 3*1.3856 = 4.1568.

Therefore,

The probability that X is within three standard deviations of the mean is approximately 1.

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There was a significant reduction in peoples over estimation of the line length, t = XXX with df = XXX, p < 0.01 two-tailed, with M = 10%, n = 25, s = 15%, Cohen's d = XXX , M = 10%, 95% CI [XXX, XXX].

8a. A. was
8b. B. t = XXX with df = XXX
8c. A. < 0.01 two-tailed
8d. C. M = 10%, n = 25, s = 15%, Cohen's d = XXX , M = 10%, 95% CI [XXX, XXX].

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Answer:

$7200

step by step Explanation:

Cameron borrowed $18,000 at an interest rate of 10% for a period of 4 years. To calculate the interest, we can use the simple interest formula: I = P * r * t, where I is the interest, P is the principal amount, r is the interest rate, and t is the time period.

Plugging in the values, we get I = 18,000 * 0.10 * 4 = $7,200. Therefore, Cameron paid a total of $7,200 in interest over the 4-year period.

The equation of a line that is perpendicular to the line y = –23x – 7 and passes through the point (–4, 2) is y = x/23 + 50/23.

In Mathematics and Geometry, the point-slope form of a straight line can be calculated by using the following mathematical equation (formula):

y - y₁ = m(x - x₁)

Where:

Since the equation of this line is perpendicular to the line y = –23x – 7, the slope is given by;

Slope, m = -23

m₁ × m₂ = -1

-23 × m₂ = -1

m₂ = -1/-23

Slope, m₂ = 1/23

At data point (-4, 2) and a slope of 1/23, a linear equation for this line can be calculated by using the point-slope form as follows:

y - y₁ = m(x - x₁)

y - 2 = 1/23(x - (-4))  

y - 2 = 1/23(x + 4)

y = x/23 + 4/23 + 2

y = x/23 + 50/23

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The circle intersects the x-axis at the points (-5, 0) and (5, 0).

We have,

Using the Pythagorean theorem to find the radius of the circle.

So,

r = √(0-3)² + (0-4)²

r = √(9+16)

  = √25

  = 5

The equation of the circle is x² + y² = 5² = 25.

To find the points where the circle intersects the x-axis,

We substitute y = 0 in the equation of the circle and solve for x:

x² + 0² = 25

x² = 25

x = ±5

Therefore,

The circle intersects the x-axis at the points (-5, 0) and (5, 0).

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To begin, let's recall that synthetic division is a method used to divide a polynomial by a linear factor (i.e. a binomial of the form x-a, where a is a constant). The result of synthetic division is the quotient of the division, which is a polynomial of one degree less than the original polynomial.

In this case, we are given that x is a solution of a third-degree polynomial equation. This means that the polynomial can be factored as (x-r)(ax^2+bx+c), where r is the given solution and a, b, and c are constants that we need to determine.

To use synthetic division, we will divide the polynomial by x-r, where r is the given solution. The result of the division will give us the coefficients of the quadratic factor ax^2+bx+c.

Here's an example of how to do this using synthetic division:

Suppose we are given the polynomial P(x) = x^3 + 2x^2 - 5x - 6 and we know that x=2 is a solution.

1. Write the polynomial in descending order of powers of x:

P(x) = x^3 + 2x^2 - 5x - 6

2. Set up the synthetic division table with the given solution r=2:

2 | 1  2  -5  -6

3. Bring down the leading coefficient:

2 | 1  2  -5  -6
  ---
   1

4. Multiply the divisor (2) by the result in the first row, and write the product in the second row:

2 | 1  2  -5  -6
  ---
   1  2

5. Add the second row to the next coefficient in the first row, and write the sum in the third row:

2 | 1  2  -5  -6
  ---
   1  2 -3

6. Multiply the divisor by the result in the third row, and write the product in the fourth row:

2 | 1  2  -5  -6
  ---
   1  2 -3
       4

7. Add the fourth row to the next coefficient in the first row, and write the sum in the fifth row:

2 | 1  2  -5  -6
  ---
   1  2 -3
       4 -2

The final row gives us the coefficients of the quadratic factor: ax^2+bx+c = x^2 + 2x - 3. Therefore, the factorization of P(x) is

P(x) = (x-2)(x^2+2x-3).

To find the real solutions of the equation, we can use the quadratic formula or factor the quadratic further:

x^2 + 2x - 3 = (x+3)(x-1).

Therefore, the real solutions of the equation are x=2, x=-3, and x=1.

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The statement that is true about the 2 functions, in which the relationship between the x and y-values in the table of values for both functions is a linear relationship is that The y-intercept of the graph of A is equal to the y-intercept of the graph of B

The equation representing the relationship in function A in point-slope form is therefore;

y - 12 = 6·(x - 2)

y - 12 = 6·x - 12

y = 6·x - 12 + 12 = 6·x

The equation in slope-intercept form, y = m·x + c, where c is the y-intercept is therefore; y = 6·x

The true statement is therefore; The y-intercept of the graph of A is less than the y-intercept of the graph of B

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Answer:

b

Step-by-step explanation:

trust ne

We have shown that (10101) times (10011) in GF(2^5) with [tex](x^5 + x^4 + x^3 + x^2+ 1)[/tex] as the modulus is equal to (101111) in binary or [tex]x^4 + x^2 + x + 1[/tex] in polynomial form.

To multiply (10101) by (10011) in GF [tex](2^5)[/tex] with [tex](x^5 + x^4 + x^3 + x^2+ 1)[/tex] as the modulus, we first need to write these polynomials as binary numbers:

[tex](10101) = 1x^4 + 0x^3 + 1x^2 + 0x + 1 = 16 + 4 + 1 = (21)_10 = (10101)_2[/tex]

[tex](10011) = 1x^4 + 0x^3 + 0x^2 + 1x + 1 = 16 + 2 + 1 = (19)_10 = (10011)_2[/tex]

We will use long multiplication to multiply these polynomials in GF[tex](2^5)[/tex], as shown below:

    1 0 1 0 1   <-- (10101)

  x 1 0 0 1 1   <-- (10011)

  ------------

    1 0 1 0 1   <-- Step 1: Multiply by 1

1 0 1 0 1      <-- Step 2: Multiply by x and shift left

------------

1 0 0 1 0 1    <-- Step 3: Add steps 1 and 2

1 0 0 1 0 <-- Step 4: Multiply by x and shift left

1 0 1 1 1 1 <-- Step 5: Add steps 3 and 4

Now, we have the product (101111)_2, which corresponds to the polynomial [tex]1x^4 + 0x^3 + 1x^2 + 1x + 1 = x^4 + x^2 + x + 1[/tex] in GF[tex](2^5)[/tex] with [tex](x^5 + x^4 + x^3 + x^2+ 1)[/tex] as the modulus. We can verify that this polynomial is indeed in GF(2^5) with modulus [tex](x^5 + x^4 + x^3 + x^2+ 1)[/tex] by noting that all of its coefficients are either 0 or 1, and none of its terms have degree greater than 4. Additionally, we can check that it satisfies the modulus:

[tex]x^4 + x^2 + x + 1 = (x^4 + x^3 + x^2 + x) + (x^3 + 1)[/tex]

[tex]= x(x^3 + x^2 + x + 1) + (x^3 + 1)[/tex]

[tex]= x(x^3 + x^2 + x + 1) + (x^3 + x^2 + x + 1)[/tex]

(since [tex]x^3 + x^2 + x + 1 = 0[/tex] in GF[tex](2^5))[/tex]

[tex]= (x+1)(x^3 + x^2 + x + 1)[/tex]

Therefore, we have shown that (10101) times (10011) in GF(2^5) with [tex](x^5 + x^4 + x^3 + x^2+ 1)[/tex] as the modulus is equal to (101111) in binary or [tex]x^4 + x^2 + x + 1[/tex] in polynomial form.

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The probability that between 124 and 125​, inclusive are on time is approximately 0.0655.

Given:

The probability of a flight arriving on time is 0.88

Number of flights selected randomly = 145

Let X be the number of flights arriving on time.

(a) P(exactly 128 flights are on time)

Using the normal approximation to the binomial distribution, we have:

Mean, µ = np = 145 × 0.88 = 127.6

Standard deviation, σ = sqrt(np(1-p)) = sqrt(145 × 0.88 × 0.12) = 3.238

P(X = 128) can be approximated using the standard normal distribution:

z = (128 - µ) / σ = (128 - 127.6) / 3.238 = 0.1234

P(X = 128) ≈ P(z = 0.1234) = 0.4511

Therefore, the probability that exactly 128 flights are on time is approximately 0.4511.

(b) P(at least 128 flights are on time)

P(X ≥ 128) can be approximated as:

z = (128 - µ) / σ = (128 - 127.6) / 3.238 = 0.1234

P(X ≥ 128) ≈ P(z ≥ 0.1234) = 0.4515

Therefore, the probability that at least 128 flights are on time is approximately 0.4515.

(c) P(fewer than 124 flights are on time)

P(X < 124) can be approximated as:

z = (124 - µ) / σ = (124 - 127.6) / 3.238 = -1.1154

P(X < 124) ≈ P(z < -1.1154) = 0.1326

Therefore, the probability that fewer than 124 flights are on time is approximately 0.1326.

(d) P(between 124 and 125​, inclusive are on time)

P(124 ≤ X ≤ 125) can be approximated as:

z1 = (124 - µ) / σ = (124 - 127.6) / 3.238 = -1.1154

z2 = (125 - µ) / σ = (125 - 127.6) / 3.238 = -0.7388

P(124 ≤ X ≤ 125) ≈ P(-1.1154 ≤ z ≤ -0.7388) = P(z ≤ -0.7388) - P(z < -1.1154)

P(124 ≤ X ≤ 125) ≈ 0.1981 - 0.1326 = 0.0655

Therefore, the probability that between 124 and 125​, inclusive are on time is approximately 0.0655.

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Mean- 6

Median- 6

Mode- 6 and 8

By the principle of mathematical induction, we can conclude that n! > 2ⁿ for all n ∈ Z≥4.

The art of demonstrating a claim, theorem, or formula that is regarded as true for each and every natural number n is known as proof.

We can prove by mathematical induction that n! > 2ⁿ for all n ∈ Z≥4.

First, we will prove the base case n = 4:

4! = 4 x 3 x 2 x 1 = 24

2⁴ = 16

Since 24 > 16, the base case is true.

Next, we assume that the inequality is true for some arbitrary k ≥ 4:

k! > [tex]2^k[/tex]

To complete the induction step, we must prove that the inequality is also true for k + 1:

(k+1)! = (k+1) x k!

(k+1)! > (k+1) x [tex]2^k[/tex]    (by the induction hypothesis)

(k+1)! > 2 x [tex]2^k[/tex]

(k+1)! > [tex]2^{(k+1)[/tex]

Since the inequality is true for k+1, this completes the induction step.

Therefore, by the principle of mathematical induction, we can conclude that n! > 2ⁿ for all n ∈ Z≥4.

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The value of x in the rectangular prism is 9 inches.

The height of the rectangular prism can be found as follows:

The volume of the rectangular prism is 153 inches cube.

Therefore,

volume of the rectangular prism = lwh

where

Therefore,

volume of the rectangular prism = 8.5 × 2 × x

153 = 17x

divide both sides by 17

x = 153 / 17

x = 9 inches

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Renting 0 trucks the Marginal Revenue (MR) = Not applicable, and Average Revenue (AR) = Not applicable. Renting 1 truck the Marginal Revenue (MR) = $P (since it's the additional revenue gained from renting 1 truck), Average Revenue (AR) = Total Revenue / Quantity = P / 1 = $P.

Renting 2 trucks Marginal Revenue (MR) = ($2P - $P) = $P (since it's the additional revenue gained from renting the second truck), Average Revenue (AR) = Total Revenue / Quantity = 2P / 2 = $P. Renting 3 trucks Marginal Revenue (MR) = ($3P - $2P) = $P (since it's the additional revenue gained from renting the third truck), Average Revenue (AR) = Total Revenue / Quantity = 3P / 3 = $P.

To help you with your question, we need to know the rental price per truck and the costs associated with renting these trucks. Since this information is not provided, I will assume a rental price of P dollars per truck. Based on this assumption, we can calculate total, marginal, and average revenue for Sendit when renting 0, 1, 2, or 3 trucks during the move-in week.

1. Renting 0 trucks:
Total Revenue (TR) = 0 * P = $0
Marginal Revenue (MR) = Not applicable
Average Revenue (AR) = Not applicable

2. Renting 1 truck:
Total Revenue (TR) = 1 * P = $P
Marginal Revenue (MR) = $P (since it's the additional revenue gained from renting 1 truck)
Average Revenue (AR) = Total Revenue / Quantity = P / 1 = $P

3. Renting 2 trucks:
Total Revenue (TR) = 2 * P = $2P
Marginal Revenue (MR) = ($2P - $P) = $P (since it's the additional revenue gained from renting the second truck)
Average Revenue (AR) = Total Revenue / Quantity = 2P / 2 = $P

4. Renting 3 trucks:
Total Revenue (TR) = 3 * P = $3P
Marginal Revenue (MR) = ($3P - $2P) = $P (since it's the additional revenue gained from renting the third truck)
Average Revenue (AR) = Total Revenue / Quantity = 3P / 3 = $P

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The number of ways there are for her to select the starting line-up is 68,640 ways.

1. Center: There are 4 players who can play this position, so there are 4 choices.
2. Right Forward: Since one player has been selected as Center, there are now 13 players remaining. So, there are 13 choices for this position.
3. Left Forward: After selecting the Center and Right Forward, 12 players remain, resulting in 12 choices for this position.
4. Right Guard: With three players already chosen, there are 11 players left to choose from, giving us 11 choices.
5. Left Guard: Finally, after selecting players for the other four positions, 10 players remain, providing 10 choices for this position.

Now, we can calculate the total number of ways to select the starting line-up using the counting principle by multiplying the number of choices for each position:

4 (Center) × 13 (Right Forward) × 12 (Left Forward) × 11 (Right Guard) × 10 (Left Guard) = 68,640 ways

So, there are 68,640 ways for the coach to select the starting line-up.

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We can state the critical values for the rejection rule as follows:

test statistic ≤ -1.645 (left-tailed test)

test statistic ≥ 1.645 (right-tailed test)

The sample mean can be calculated by adding up all the weekly pays and dividing by the sample size:

sample mean = (687.73 + 543.15 + ... + 703.06) / 50 = 638.55 (rounded to two decimal places)

To test whether the average weekly earnings for workers who have received a high school diploma is significantly greater than average weekly earnings for workers who have not received a high school diploma, we can perform a two-sample t-test assuming equal variances. The null hypothesis is that there is no difference in the means of the two groups, and the alternative hypothesis is that the mean for the high school diploma group is greater than the mean for the non-high school diploma group.

Using a calculator or software, we can calculate the test statistic and p-value. Assuming a two-tailed test and a significance level of 0.05, the critical values for the rejection rule are -1.96 and 1.96.

test statistic = 3.196 (rounded to three decimal places)

p-value = 0.0012 (rounded to four decimal places)

Since the p-value (0.0012) is less than the significance level (0.05), we reject the null hypothesis and conclude that the average weekly earnings for workers who have received a high school diploma is significantly greater than average weekly earnings for workers who have not received a high school diploma.

For a one-tailed test with α = 0.05, the critical value would be 1.645. The rejection rule would be: if the test statistic is greater than 1.645, reject the null hypothesis. Therefore, we can state the critical values for the rejection rule as follows:

test statistic ≤ -1.645 (left-tailed test)

test statistic ≥ 1.645 (right-tailed test)

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Since the calculated chi-square value (3.91) is less than the critical value (7.82) and the significance level is 0.05, the null hypothesis cannot be rejected.

The null hypothesis in a chi-square goodness-of-fit test for independent assortment is that the observed data fits the expected data under the assumption of independent assortment. Therefore, we conclude that there is no significant difference between the observed and expected data under the assumption of independent assortment at a significance level of p=0.05.

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The 80% confidence interval for the mean waste recycled per person per day for the population of Montana is given as follows:

(1.9, 2.3).

The t-distribution is used when the standard deviation for the population is not known, and the bounds of the confidence interval are given according to the following rule:

[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]

In which the variables of the equation are presented as follows:

The critical value, using a t-distribution calculator, for a two-tailed 80% confidence interval, with 18 - 1 = 17 df, is t = 1.33.

The parameters are given as follows:

[tex]\overline{x} = 2.1, s = 0.74, n = 18[/tex]

The lower bound of the interval is given as follows:

2.1 - 1.33 x 0.74/sqrt(18) = 1.9.

The upper bound of the interval is given as follows:

2.1 + 1.33 x 0.74/sqrt(18) = 2.3.

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Answer:

The answer is D.) A random sample may represent the population.

Step-by-step explanation:

Collecting samples from anyone at random increases the chances of representing the overall people because that's where they're getting it from. But there is still a chance it will not represent everyone or the general population.

Answer:B) A random sample may represent the population is your best answer.

Step-by-step explanation:

this is what i got for this one. have a good day everyone. :)

The equation of the parabola is given as follows:

y = -28/19600(x - 140)² + 28.

The distances from the center for a height of 13 meters are given as follows:

37.53 m and 242.47 m.

The equation of a parabola of vertex (h,k) is given by the equation presented as follows:

y = a(x - h)² + k.

In which a is the leading coefficient.

It has a span of 280 meters, hence the x-coordinate of the vertex is given as follows:

x = 280/2

x = 140 -> h = 140.

The maximum height is of 28 meters, hence the y-coordinate of the vertex is given as follows:

y = 28 -> k = 28.

Hence the equation is:

y = a(x - 140)² + 28.

When x = 0, y = 0, hence the leading coefficient a is obtained as follows:

19600a = -28

a = -28/19600

Hence:

y = -28/19600(x - 140)² + 28.

The distance from the center at which the height is 13 meters is obtained as follows:

13 = -28/19600(x - 140)² + 28.

28/19600(x - 140)² = 15

(x - 140)² = 15 x 19600/28

(x - 140)² = 10500.

Hence the distances are obtained as follows:

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The solution to the differential equation is y(t) = 3cosh(3t) + 2sin(4t).

To solve this differential equation using Laplace transform, we first apply the transform to both sides of the equation:

L[day + 2dy/dt + y] = L[sinh(3t) - 5cosh(3t)]

Using the properties of Laplace transform and the derivative property, we get:

sY(s) - y(0) + 2[sY(s) - y(0)]/dt + Y(s) = 3/(s^2 - 9) - 5s/(s^2 - 9)

Substituting the initial conditions y(0) = -2 and y'(0) = 5, and simplifying the expression, we get:

Y(s) = (3s - 19)/(s^3 - 2s^2 - 3s + 18)

Now, we need to find the inverse Laplace transform of Y(s) to obtain the solution y(t). This can be done using partial fraction decomposition, which gives:

Y(s) = -1/(s - 3) + 4/(s + 2) + 2/(s - 3)^2

Taking the inverse Laplace transform of each term using the Laplace transform table, we get:

y(t) = -e^(3t) + 4e^(-2t) + 2te^(3t)

Therefore, the solution to the differential equation is y(t) = -e^(3t) + 4e^(-2t) + 2te^(3t).

To solve this differential equation using Laplace transform, we first apply the transform to both sides of the equation:

L[day/dt^2 - 4y - 5y] = L[e^3 + sin(4t)]

Using the properties of Laplace transform, we get:

s^2Y(s) - sy(0) - y'(0) - 4Y(s) - 5Y(s) = 3/(s - 3) + 4/(s^2 + 16)

Substituting the initial conditions y(0) = 0 and y'(0) = 0, and simplifying the expression, we get:

s^2Y(s) - 9Y(s) = 3/(s - 3) + 4/(s^2 + 16)

Using partial fraction decomposition, we get:

Y(s) = (3s - 9)/(s^2 - 9) + (4s)/(s^2 + 16)

Taking the inverse Laplace transform of each term using the Laplace transform table, we get:

y(t) = 3cosh(3t) + 2sin(4t)

Therefore, the solution to the differential equation is y(t) = 3cosh(3t) + 2sin(4t).

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