Calculate the size of angle x.
x
240°
x
x

The fundamental set of solutions of the given system of differential equations x' =(1 2 0, -3 -1 3, 3 2 -2) is to be identified from the given options.

The correct answer is option (a) x1 = e^-2t(2 -3 3), X2 = e^-t(1 -1 1), X3 = e^t(1 0 1).

To verify this, we can calculate the Wronskian of the three solutions and show that it is non-zero, which confirms that they form a fundamental set of solutions. Another way to check is to substitute the solutions into the differential equation and verify that they satisfy it. In this case, both methods give us the same result - the solutions satisfy the differential equation and are linearly independent, hence form a fundamental set of solutions. Therefore, the correct answer is (a).

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The equations that are true for x= –2 and x = 2 is x²-4=0 and 4×2=16

x= -2, x=2: x²-4=0

x²= 0 (multiplying the value of x with itself)

Since x² means multiplying the value of x with itself and 4 times itself equals 16, the next value that equals x = –2 and x = 2 is 4×2=16

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a. the horizontal sum of all individual firm supply schedules at this output level. b. output level, each firm will earn zero economic profit (normal profit).

a) In the long-run, each firm will produce 20 units of neckties. The industry supply schedule will be the horizontal sum of all individual firm supply schedules at this output level.

b) The long-run equilibrium price (P*) is $20 per unit, with a total industry output (Q*) of 1,000 units. Each firm will produce 20 units of neckties, and the number of firms in the industry will be 50. At this output level, each firm will earn zero economic profit (normal profit).

c) The short-run average cost curve can be found by dividing the short-run total cost by output. Thus, the short-run average cost curve is SAC = 0.5q - 10 + 200/q. The short-run marginal cost curve is SMC = q - 10. Short-run average cost reaches a minimum at an output level of 20 units.

d) The short-run supply curve for each firm is the portion of the marginal cost curve above the average variable cost curve. The industry short-run supply curve is the horizontal sum of all individual firm supply curves.

e) With the new demand curve, the short-run equilibrium price (P*) is $30 per unit. The total industry output (Q*) is 1,250 units, with each firm producing 25 units of neckties.

f) In the short run, the industry short-run supply curve will shift upwards, resulting in a higher equilibrium price and output level. The new short-run equilibrium price (P*) will be higher than $20 per unit and the new total industry output (Q*) will be higher than 1,000 units.

g) In the long run, new firms will enter the industry, causing the supply curve to shift to the right until price falls back to the minimum long-run average cost of $10 per unit. At the new long-run equilibrium, each firm will produce 20 units of neckties, the industry output (Q*) will increase, and the price (P*) will fall back to $20 per unit.

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Two consecutive integers such that the square of the larger integer is 19 more than 9 times the smaller integer are 9 and 10

Let x be the smaller integer, then the larger integer is x + 1. According to the problem, we can set up an equation:

(x + 1)^2 = 9x + 19

Expanding the left side and simplifying, we get:

x^2 + 2x + 1 = 9x + 19

Bringing all the terms to one side, we get:

x^2 - 7x - 18 = 0

Factorizing, we get:

(x - 9)(x + 2) = 0

So, x = 9 or x = -2. Since we are looking for consecutive integers, we can discard the negative solution. Therefore, the smaller integer is 9 and the larger integer is 10. We can verify that this solution satisfies the original equation:

10^2 = 100 = 9(9) + 19 = 82

So, the two consecutive integers are 9 and 10.

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(a) The researcher should obtain a sample size of 1,068 households to estimate the proportion of households with broadband internet access within 0.03 with 99% confidence, assuming a prior estimate of 0.635 from 2009.

(b) If the researcher does not use any prior estimates, she can use a conservative estimate of 0.5 for the proportion of households with broadband internet access, as this value maximizes the sample size required for a given level of precision and confidence. With this assumption, the researcher should obtain a sample size of 1,068 households to estimate the proportion of households with broadband internet access within 0.03 with 99% confidence. It is important to note that if the true proportion is significantly different from 0.5, the required sample size may be higher or lower than this estimate. Additionally, the researcher should consider other factors such as the cost and feasibility of obtaining a sample of this size.

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The ratio of red apples to all apples in the basket is 3:11, whereas all apples to green is 11:8

Total number of green apples = 8

Total number of red apples = 3

Calculating the total number of apples -

Total number of green apples + Total number of red apples

= 8 + 3

= 11

Calculating the ratio of red apples to all apples in the basket -

= Total number of red apples / Total number of apples

= 3/11

Thus, for every 11 apples in the basket, 3 of them are red.

Calculating the ratio of all apples in the basket to green apples -

Total number of apples / Total number of green apples

= 11/8.

Thus, for every 8 green apples in the basket, there are a total of 11 apples in the basket.

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Answer:

Step-by-step explanation:

To find the probability that a student plays both basketball and baseball, we need to determine the number of students who play both sports and divide it by the total number of students in the class.

Given:

Total number of students (n) = 21

Number of students who play basketball (B) = 10

Number of students who play baseball (A) = 9

Number of students who play neither sport = 8

Let's calculate the number of students who play both basketball and baseball (B ∩ A):

Number of students who play both sports (B ∩ A) = Number of students who play basketball (B) + Number of students who play baseball (A) - Total number of students (n) + Number of students who play neither sport

B ∩ A = B + A - n + Neither

B ∩ A = 10 + 9 - 21 + 8

B ∩ A = 6

The number of students who play both basketball and baseball is 6.

Now, we can calculate the probability:

Probability of playing both basketball and baseball = Number of students who play both sports (B ∩ A) / Total number of students (n)

Probability = 6 / 21

Probability = 2 / 7

Therefore, the probability that a student chosen randomly from the class plays both basketball and baseball is 2/7.

1. Average rate of change is 17.5 and 22.5 respectively.

2. The Strain B is growing faster.

To get average rate of change for each strain from week 0 to 4, we will use the formula: [tex]Average rate of change = (change in cases) / (change in weeks)[/tex]

For Strain A:

Change in cases = 85 - 15 = 70

Change in weeks = 4 - 0 = 4

Average rate of change = 70/4

Average rate of change = 17.5

For Strain B:

Change in cases = 115 - 25 = 90

Change in weeks = 4 - 0 = 4

Average rate of change = 90/4

Average rate of change = 22.5

The Strain B is growing faster with an average rate of change of 22.5 cases per week compared to Strain A with an average rate of change of 17.5 cases per week.

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The acute angle between the lines is approximately 37 degrees.

To find the acute angle between the lines given by the equations 2x - y = 3 and 6x + y = 9, we can compare the slopes of the lines.

The slope-intercept form of a line is y = mx + b, where m is the slope. By rearranging the given equations into this form, we can determine the slopes.

For the first equation, 2x - y = 3, we can rewrite it as y = 2x - 3. The slope of this line is 2.

For the second equation, 6x + y = 9, we can rewrite it as y = -6x + 9. The slope of this line is -6.

To find the acute angle between the lines, we can use the formula:

angle = arctan(|m1 - m2| / (1 + m1 * m2))

Plugging in the slopes:

angle = arctan(|2 - (-6)| / (1 + 2 * (-6)))

Simplifying the expression:

angle = arctan(8 / (-11))

Using a calculator or trigonometric tables, we can find:

angle ≈ -37.15 degrees

Since we are looking for the acute angle, we take the absolute value of the result:

acute angle ≈ 37 degrees

Therefore, the acute angle between the lines is approximately 37 degrees.

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The correct answer is option B, which is 1/(x+4). To see why, first factor the denominator of the given expression:

x^2 - 16x + 64 = (x - 8)(x - 8) = (x - 8)^2

Now, we can rewrite the original expression as:

(x - 8)^2 / [(x - 8)(x + 4)]

Canceling the common factor (x - 8), we get:

(x - 8) / (x + 4)

This is equivalent to 1/(x+4) since (x - 8) / (x + 4) = (x + 4 - 12) / (x + 4) = 1 - 12 / (x + 4) = 1 - 3 / (x + 4/3). As x approaches infinity, 3/(x+4/3) approaches 0, so 1 - 3 / (x + 4/3) approaches 1. Thus, the expression is equivalent to 1/(x+4) for any value of x except x = -4.

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Answer:

Step-by-step explanation:

Let x - be the number of sweets in small packs

y - be the number of sweets in big packs

Therefore, we have:

4x + 3y = 175 (1)

5x + 2y = 154 (2)

Now, we find the difference between (1) & (2) is:

y-x = 21. Thus, y = 21+x

Now we substitute the value of y = 21+x to any of the two statements, we have 4x + 3(21+x) = 175 => 4x + 63 + 3x = 175.

Hence, 7x = 175 - 63 = 112 or simply, x=16.

Now, finding the value of y:

5(16) + 2y = 154

80 + 2y = 154

2y = 154-80

2y = 74

y = 37.

Therefore, there are 16 sweets in each small pack and 37 sweets in each big pack.

The truckers average speed in kilometers per hour would be = 80.9 km/hr

To calculate the average speed of the trucker the formula for speed should be used and this is given below;

Speed = Distance/ time

Distance = 849 km

Time = 10 hours 30 minutes= 10.5 hours

Speed = 849/10.5 = 80.9km/hr

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Answer:

$300

Step-by-step explanation:

The area of the area would be 10 x 12 = 120 square feet.

120 sq ft x 2.50 per sq ft  = $300.

It would cost $300 to carpet this area.

If the test procedure with α = 0.003 is used, the n that is necessary for β(70) = 0.01 would be 59.

To find the sample size n necessary to ensure that β ( 70 ) = 0. 01, we can use the following formula:

n = ( zα + zβ ) ²  * σ ² / Δ ²

In this case, we have:

zα = z (0. 003) = 2. 576

zβ = z (0.01 ) = 2. 326

σ = 6

Δ = 70 - 74

= -4

This means that the value of n is:

= (2. 576 + 2. 326) ²  * 6 ² / ( - 4) ²

= 58. 28

= 59

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A sampling plan is a vital component in conducting statistical process control smoothly, as it provides structure, reduces variability, identifies critical parameters, guides data analysis, and enhances decision-making.

1. Defining the sample size and frequency: A sampling plan establishes the number of items to be collected and the intervals at which they will be collected. This ensures a consistent and representative sample, making the statistical process more reliable and efficient.

2. Reducing variability: By specifying the method of sample selection, a sampling plan helps minimize the potential for biased or non-representative samples. This results in better control over the process and more accurate insights into the system's performance.

3. Identifying critical parameters: A sampling plan helps identify the key characteristics of a process that need to be monitored and controlled. This enables the focus on essential aspects of the process, ensuring optimal control and improvement efforts.

4. Guiding data analysis: A well-established sampling plan provides a structure for data collection, which can be used to perform statistical process control. It aids in data organization and interpretation, making it easier to detect trends, patterns, and potential issues.

5. Enhancing decision-making: With a sampling plan in place, statistical process control results become more trustworthy and actionable. This allows for better-informed decisions related to process adjustments and quality improvement initiatives.

In summary, a sampling plan is a vital component in conducting statistical process control smoothly, as it provides structure, reduces variability, identifies critical parameters, guides data analysis, and enhances decision-making.

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Answer:

Answer: 3.90125×10⁻³

Step-by-step explanation:

The standard deviation of the sample is approximately 0.0180.

To find the standard deviation of a sample with a sample size (n) and proportion (p), we can use the formula:

Standard deviation (σ) = √(p(1-p)/n)

Given that n = 200 and p = 0.07, we can substitute these values into the formula:

σ = √(0.07(1-0.07)/200)

σ = √(0.07(0.93)/200)

σ = √(0.0651/200)

σ ≈ √0.0003255

σ ≈ 0.01803

Rounding to four decimal places, the standard deviation of the sample is approximately 0.0180.

Comparing this result with the given options, none of them match exactly. However, the option closest to the calculated standard deviation is b. 0.0324. It is important to note that this option is not an exact match and may be considered an error or an approximation. The actual standard deviation based on the given values is approximately 0.0180.

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The solution is: The surface area of the cuboid is: 1900 cm².

Here, we have,

We can use the given ratios and volume to find the scale factor for the dimensions. Knowing the dimensions, we can compute the surface area using the formula for a cuboid.

dimensions

Let k represent the scale factor. Then the actual dimensions will be 5k, 4k, and 2k. The actual volume will be ...

 V = LWH

 5000 cm³ = (5k)(4k)(2k) = 40k³

 k³ = (5000 cm³)/40 = 125 cm³

 k = ∛(125 cm³) = 5 cm

The cuboid dimensions are 5(5 cm) = 25 cm, 4(5 cm) = 20 cm, and 2(5 cm) = 10 cm.

area

The surface area of the cuboid can be computed from ...

 A = 2(LW +H(L +W))

 A = 2((25 cm)(20 cm) +(10 cm)(25 +20 cm))

 A = 2(500 cm² +(10 cm)(45 cm)) = 2(950 cm²) = 1900 cm²

The surface area of the cuboid is 1900 cm².

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complete question:

the length, breadth and height of a cuboid are in the ratio 5:4:2 if the volume of cuboid is 5000 cm,then what will be the surface of the area of the cuboid​

The function is Neither one-to-one nor onto. An example of a function that is one-to-one but not onto is f(x) = x + 1, where the domain is all integers and the target is all positive integers.

The function h is neither one-to-one nor onto.

It is not one-to-one because for example, h({a}) = h({b}) since h({a}) = {a, b} and h({b}) = {a, b}.

It is not onto because {b, c} is not in the range of h since h(X) always contains a but {b, c} does not contain a.

One example of a function with the given properties is f(x) = x + 1.

It is one-to-one because for any distinct integers x and y, f(x) = x + 1 and f(y) = y + 1 are different since x and y are different.

It is not onto because the target set of f only includes positive integers, but there is no integer x such that f(x) = 1.

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We have that about 50% of percentage of the students had scores between 485 and 695.

Option B is correct.

A Box-and-Whisker Plot  is  described as a method for graphically demonstrating the locality, spread and skewness groups of numerical data through their quartiles.

The box in the plot represents the interquartile range, therefore  the percentage of students who scored between the lower quartile and the upper quartile of the distribution, are those  between the edges of the box.

We take a look at the percentile ranks associated with those scores. and find the  estimate of percentile ranks by drawing a horizontal line at the score values and then reading the corresponding percentile ranks off the y-axis.

With reference from the plot, a score of 485 appears to be at or below the 50th percentile, while a score of 695 appears to be around the 100th percentile.

We then have that the percentage of students with scores between 485 and 695 is likely to be between 100% - 50% = 50%.

The interquartile range represents the middle 50% of the data and the box covers this range.

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The height of the smaller cone is 20/7, the correct option is B.

We are given that;

The two cones

Now,

To find the height of the smaller cone, you need to use the similarity ratio of the cones. Similar cones have proportional dimensions, so you can set up a proportion between the corresponding heights and radii. You can write your solution as:

h/7 = 20/10 h = 20/10 x 7 h = 14

Therefore, by the proportion the answer will be 20/7.

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The total distance Trevor traveled is 352 meters.

We have,

Trevor is traveling from his home to school.

He first walks 58.0 meters in one direction, but then he forgets his lunch and has to turn around and walk back the same distance.

This means he has walked a total distance of 58.0 m + 58.0 m = 116.0 m.

Now,

After he retrieves his lunch, he continues walking in the original direction for an additional 236 meters.

So, the total distance Trevor traveled.

= 116.0 m + 236 m = 352.0 m.

Thus,

The total distance Trevor traveled is 352 meters.

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The probability that x = 22.56 is the 1.64

The probability formula defines the likelihood of the happening of an event. It is the ratio of favorable outcomes to the total favorable outcomes. The probability formula can be expressed as,

P(A) = Number of favorable outcomes of A / Total number of possible outcomes.

We must standardize the Random Variable X with the standardized Normal distribution Z variable using the relationship:

[tex]Z =\frac{X-\mu}{\sigma}[/tex]

We have the information from the question:

Mean ([tex]\mu[/tex]) = 16

Standard deviation ([tex]\sigma[/tex]) = 4

To find the probability that x equals 22.56.

P(X= 22.56) = [tex]P(\frac{22.56-16}{4} )[/tex]

                   = [tex]P(\frac{6.56}{4} )[/tex]

                   = P(1.64)

Hence, The probability that x = 22.56 is the 1.64

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The cubic function only has one real zero, which is x = -2 - √5/3  = -2.75

Here we want to find the zeros of the cubic function:

y = 27(x + 2)³ + 5

The zeros of a function are the values of x such that the outcome is y, then we need to solve the equation:

0 =  27(x + 2)³ + 5

-5 =  27(x + 2)³

-5/27 = (x + 2)³

∛(-5/27) = x + 2

-√5/3 = x + 2

-2 - √5/3 = x

That is the only zero of the function (with a multiplicity of 3).

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The standard form of the equation is 2x^2 + 24y - 35 = 0. To find the standard form of the equation of a parabola with vertex at the origin, we need to use the formula 4p(x^2 + y^2) = (x - h)^2 + (y - k)^2, where (h,k) is the vertex and p is the distance from the vertex to the focus (or directrix, depending on the given information).

For the first characteristic, we have a focus at (-9,0), which is to the left of the vertex at the origin. This means that p = 9 (the distance from the vertex to the focus). Substituting into the formula, we get:
4(9)(x^2 + y^2) = (x - 0)^2 + (y - 0)^2
36x^2 + 36y^2 = x^2 + y^2
35x^2 + 35y^2 = 0
So the standard form of the equation is 35x^2 + 35y^2 = 0. For the second characteristic, we have a focus at (0,1/6), which is above the vertex at the origin. This means that p = 1/6 (the distance from the vertex to the focus). We also know that the directrix is a horizontal line 2 units below the vertex. This means that the equation of the directrix is y = -2. Using the formula and the distance formula between a point and a line, we can write:
4(1/6)(x^2 + y^2) = (y - 0)^2 - 2^2
2x^2 + 24y - 35 = 0

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The theater has a total of 945 seats.

To determine the number of seats in the theater, we need to calculate the sum of seats in each row. The first row has 25 seats, and each subsequent row increases by one seat. Since there are 35 rows in total, we can calculate the sum of an arithmetic series to find the total number of seats.

The formula for the sum of an arithmetic series is Sn = (n/2) * (a1 + an), where n is the number of terms, a1 is the first term, and an is the last term. In this case, n = 35 (number of rows), a1 = 25 (number of seats in the first row), and an = a1 + (n - 1) = 25 + (35 - 1) = 25 + 34 = 59 (number of seats in the last row). Plugging these values into the formula, we get Sn = (35/2) * (25 + 59) = 17.5 * 84 = 1470. Therefore, the theater has a total of 945 seats.


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The estimated probability of the total sum being at least 360 is approximately 8%.

To estimate the probability using a normal approximation with a continuity correction, we first need to find the mean and standard deviation of the sum of the numbers on the 50 dice.
For a single 12-sided die, the mean is (1+2+...+12)/12 = 6.5. For 50 dice, the mean is 50 × 6.5 = 325. The variance for one die is [(1-6.5)²+(2-6.5)²+...+(12-6.5)²]/12 = 11.92. For 50 dice, the variance is 50 × 11.92 = 596, and the standard deviation is √596 ≈ 24.4.
Now, we'll use the normal approximation with a continuity correction to estimate the probability that the sum of the numbers is at least 360. First, find the z-score:
z = (X - μ + 0.5) / σ = (360 - 325 + 0.5) / 24.4 ≈ 1.42
Using a z-table or calculator, the probability of obtaining a z-score greater than 1.42 is approximately 0.0778 or 7.78%. The closest percentage in the options provided is 8%, which corresponds to option g. Therefore, the estimated probability of the total sum being at least 360 is approximately 8%.

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we know the radius has a diameter of 26 cm, so its radius must be half that, or 13 cm.

[tex]\textit{area of a circle}\\\\ A=\pi r^2 ~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=13 \end{cases}\implies A=\pi (13)^2 \\\\\\ A=(3.14)(13)^2\implies A=530.66~cm^2 \\\\[-0.35em] ~\dotfill\\\\ \textit{circumference of a circle}\\\\ C=2\pi r ~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=13 \end{cases}\implies C=2\pi 13 \\\\\\ C=2(3.14)(13)\implies C=81.64~cm[/tex]

Check the picture below.

The probability of winning this lottery game is 1/1,000 or 0.001, which equates to a 0.1% chance.

In this lottery game, players select a three-digit number, ranging from 000 to 999. Since repetition is allowed, each digit can be any number between 0 and 9, giving a total of 10 options per digit. The three digits are independent, which means that the choice of one digit does not influence the choices for the other digits. Consequently, to find the total number of possible combinations, you can use the counting principle.

The counting principle states that if there are n ways to do one thing and m ways to do another, there are n x m ways to do both. In this case, there are 10 choices for each of the three digits, so the total number of combinations is 10 x 10 x 10 = 1,000.

Players win the lottery game if their chosen three-digit number matches the winning number drawn by the game organizers. The probability of winning is determined by dividing the number of successful outcomes (1, as there's only one winning number) by the total number of possible outcomes (1,000 combinations). Hence, the probability of winning is 1/1,000 or 0.1 % chance.

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The correct answer is C) [tex]20[/tex] ≤ [tex]h[/tex]  < [tex]30[/tex]. This class interval contains the median height in the given table of plant heights.

To identify the class interval containing the median in the given table, we analyze the cumulative frequency of the height data. Cumulative frequency is the running total of frequencies as we progress from the lowest height to the highest height.

Examining the provided table, we observe the following frequencies for each class interval:

The interval [tex]0[/tex] ≤ h < [tex]10[/tex] has a frequency of [tex]1[/tex].

The interval [tex]10[/tex] ≤ h < [tex]20[/tex] has a frequency of [tex]20[/tex].

The interval [tex]20[/tex] ≤ h < [tex]30[/tex] has a frequency of [tex]30[/tex].

To find the median, we need to determine the class interval that encompasses the middle value. Since the total number of data points is [tex]30[/tex], the midpoint would be the [tex]15th[/tex] value.

Starting from the lowest class interval, we track the cumulative frequency. We see that the cumulative frequency for the interval [tex]0[/tex] ≤ h < [tex]10[/tex] is [tex]1[/tex], and it increases to [tex]20[/tex] for the interval [tex]10[/tex] ≤ h < [tex]20[/tex]. However, this cumulative frequency does not yet reach the midpoint.

Finally, for the interval [tex]20[/tex] ≤ h < [tex]30[/tex], the cumulative frequency is [tex]30[/tex], exceeding the midpoint value. This indicates that the median falls within the class interval [tex]20[/tex] ≤ h < [tex]30[/tex].

Therefore, the correct answer is C) [tex]20[/tex] ≤ h < [tex]30[/tex]. This class interval contains the median height in the given table of plant heights.

Table:

+--------------------+----------------+

| Class Interval | Frequency |

+--------------------+-----------------+

| 0 ≤ h < 10       |            1        |

| 10 ≤ h < 20     |          20       |

| 20 ≤ h < 30    |          30       |

+--------------------+-----------------+

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