Are the data shown in this line plot skewed left, skewed right, or not skewed?

Responses

skewed left
not skewed
skewed right

Answer:5/14

Step-by-step explanation:

you have 5 red and 14 total so your probability is 5/14 and 5/14 is the simplest terms

The Equation of circle is (x+3)² + (y-5)² = (√20)².

We have,

Center = (-3, 5)

Point = (1, 7)

We know the standard form of Equation of circle

(x-h)² + (y-k)² = r²

where (x, y) is any point on the circle, (h, k) is the center

So, (x+3)² + (y-5)² = r²

Put the point (1, 7) in above equation we get

(1+3)² + (7-5)² = r²

(4)² + (2)² = r²

16 + 4= r²

r= √20

Thus, the Equation of circle is

(x+3)² + (y-5)² = (√20)²

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Based on the data provided, we can calculate the slope and y-intercept of the line that fits the data points.

First, let's find the slope (m) using the formula: m = (y2 - y1) / (x2 - x1). We can use the first two data points for this calculation:

m = (27 - 32) / (10 - 5) = (-5) / 5 = -1

Now, let's find the y-intercept (b) using the formula: y = mx + b. We can use the first data point (5, 32) and the slope we found:

32 = -1 * 5 + b
32 = -5 + b
b = 37

Therefore, the correct slope is -1, and the y-intercept is 37.

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The point (1, -4) does not make the equation y = x + 5 true.

Given information:

The equation is y = x + 5.

To check if the point (1, -4) makes the equation y = x + 5 true:

we need to substitute the x and y values of the point into the equation and see if the equation is true.

y = x + 5

-4 = 1 + 5

-4 = 6

The equation is not true when we substitute the values of x = 1 and y = -4 into it.

Therefore, the point (1, -4) does not make the equation y = x + 5 true.

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The correct sample space for this random experiment is:
{BG, GB, BBG, BGB, GBB, GGB, GBG, GG} where B represents a boy and G represents a girl.

To find the correct sample space:

The first two outcomes, BG and GB, represent the couple having one boy and one girl in their first two children, respectively.

The next three outcomes, BBG, BGB, and GBB, represent the couple having two boys and a girl, a boy, a girl and a boy, and a girl and two boys, respectively, before having at least one boy and one girl.

The last three outcomes, GGB, GBG, and GG, represent the couple having two girls and a boy, a girl, a boy and two girls, and three girls, respectively, before having at least one boy and one girl

The sample space includes all possible combinations of children that the couple can have, including having only one boy or one girl, or having up to three children but stopping after they have one boy and one girl.

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It is difficult to say much more about the sample mean score.

If we know the sample mean score, which is denoted by x in your question, we can use it to make some inferences about the overall population of second-graders in that school district. However, we would need more information about the distribution of scores, such as the standard deviation or the range, to draw any conclusions about the entire population.

For example, if we assume that the distribution of scores is approximately normal, we could use the sample mean and standard deviation to calculate a confidence interval for the population mean score. This interval would give us a range of scores within which we can be reasonably confident the true population mean falls.

Without more information about the sample or the population, it is difficult to say much more about the sample mean score.

Complete question: Are you smarter than a second-grader? A random sample of 45 second-graders in a certain school district are given a standardized mathematics skills test. The sample mean score is x-54. Assume the standard deviation of test scores is o = 15. The nationwide average score on this test is 50. The school superintendent wants to know whether the second-graders in her school district have different math skills from the nationwide average. Use the a=0.05 level of significance and the P-value method with the TI-84 calculator al Part: 0/4 Part 1 of 4 State the appropriate null and alternate hypotheses.

Previous question

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the answer is b or the second one

Answer:

Step-by-step explanation:

first do Pemdas and thats it

The number of vases painted in the second week is given as follows:

39 vases.

The number of vases painted in the second week is obtained applying the proportions in the context of the problem.

They painted 225% more vases than in the first week, hence the equivalent percentage is of 100 + 225 = 325%, which is 3.25 times more vases than in the first week.

They painted 12 vases in the first week, hence the number of vases painted in the second week is given as follows:

3.25 x 12 = 39 vases.

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Based on the given information, we know that the medium-sized jet has a fuselage diameter of 3.8 meters and a loaded mass of 85,000 kilograms. The drag on the airplane is primarily due to the cylindrical shape of the fuselage, and its aerodynamic shaping gives it a drag coefficient of 0.37.

To calculate the drag force on the airplane, we can use the formula:

Drag Force = 1/2 x Density x Velocity^2 x Surface Area x Drag Coefficient

The surface area of a cylinder is given by:

Surface Area = 2 x π x (diameter/2) x length

Assuming a length of 30 meters for the fuselage, we can calculate the surface area as:

Surface Area = 2 x π x (3.8/2) x 30 = 426.43 square meters

Using the given drag coefficient of 0.37 and assuming a cruising speed of 800 km/h (or 222.22 m/s), we can calculate the drag force as:

Drag Force = 1/2 x 1.225 kg/m^3 x (222.22 m/s)^2 x 426.43 m^2 x 0.37
Drag Force = 5,641,613 Newtons

Therefore, the drag force on the medium-sized jet is approximately 5.64 million Newtons. This drag force must be overcome by the jet's engines to maintain its cruising speed.

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Based on the given information, the capacity problem seems to be infeasible or unrealistic. This is because the resource demands (ci) for week 3, 4, and 5 are significantly higher than the available resources (ri) for those weeks. This implies that there is a problem with capacity constraints and the organization may not be able to meet the demand for those weeks.

Hi there! Based on the given data for capacity (ci) and requirements (ri) over weeks 1 to 5, the capacity problem appears feasible. Here's the comparison for each week:

Week 1: ri = 75, ci = 100 (Capacity is sufficient)
Week 2: ri = 100, ci = 200 (Capacity is sufficient)
Week 3: ri = 200, ci = 300 (Capacity is sufficient)
Week 4: ri = 100, ci = 50 (Capacity is insufficient)
Week 5: ri = 250, ci = 50 (Capacity is insufficient)

The capacity problem is feasible for weeks 1 to 3, but not for weeks 4 and 5, as the capacity is insufficient to meet the requirements in those weeks.

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The probability of winning the lottery after at least n trials, we can use the formula: P(X≤ n) = 1 - (1-p)^n

Where X is the number of trials needed to win the lottery, p is the probability of winning the lottery in each trial, and n is the minimum number of trials we are considering.

To understand the above formula, we need to consider the probability of winning the lottery in each trial. Let's assume that the probability of winning the lottery in each trial is p. Therefore, the probability of not winning with the same ticket is 1-p in each trial.

Now, let's consider the random variable X, which represents the number of trials needed to win the lottery. If we want to calculate the probability of winning the lottery after at least n trials, we need to find the probability of winning in the first n trials, plus the probability of winning in the (n+1)th trial or any subsequent trial.

The probability of winning in the first n trials can be calculated using the binomial distribution formula, which is:

P(X=k) = (n choose k) * p^k * (1-p)^(n-k)

Where k is the number of trials in which we win the lottery.

However, we are interested in finding the probability of winning after at least n trials, which means that we need to consider all possible values of k greater than or equal to n. Therefore, we can use the cumulative distribution function (CDF) of the binomial distribution to calculate the probability of winning after at least n trials, which is:

P(X≥ n) = 1 - Σ P(X=k) for k=0 to n-1

This formula calculates the probability of winning the lottery in the first n-1 trials and subtracts it from 1 to get the probability of winning in the nth trial or any subsequent trial.

However, we are interested in the probability of winning after at least n trials, not just in the nth trial or any subsequent trial. Therefore, we can rewrite the above formula as:

P(X≤n) = 1 - P(X>n-1)

Which gives us the probability of winning the lottery after at least n trials, using the complement rule of probability.

Finally, substituting the binomial distribution formula in the above formula, we get:

P(X≤n) = 1 - (1-p)^n

This formula gives us the probability of winning the lottery after at least n trials, based on the probability of winning the lottery in each trial and the minimum number of trials we are considering.

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The minimum weight of the middle 95% of the players is 185.4 pounds.

To find the minimum weight of the middle 95% of the players, we need to find the weight that separates the bottom 2.5% of the distribution from the top 2.5%.

We can use the z-score formula:

z = (x - μ) / σ

Where:
x = the weight we're looking for
μ = the mean weight of 205 pounds
σ = the standard deviation of 10 pounds

To find the z-score that corresponds to the bottom 2.5%, we can use a z-table or calculator and look up the z-score that has an area of 0.025 to its left. This value is -1.96.

Plugging this into the z-score formula:

-1.96 = (x - 205) / 10

Solving for x:

x = (-1.96 * 10) + 205

x = 185.4

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Answer:

Step-by-step explanation:

2 and 1/2

Answer:

2 1/2 or 2.5

Step-by-step explanation:

1. Rewrite:

1 1/2 + x = 4

2. Subtract 1 1/2 from both sides:

4 - 1 1/2 = 2 1/2

x = 2 1/2 or 2.5

Answer:

x = 45

Step-by-step explanation:

y is the score on the test. To find the amount of time to study he needs to get a score of 90 on his next test, we need to find x ("y=1/5x+81 models his score after studying for x minutes.")

Initial equation
[tex]y = \frac{1}{5}x + 81[/tex]

Replace y with 90
[tex]90 = \frac{1}{5}x + 81[/tex]

Find for x
[tex]90 - 81 = \frac{1}{5}x[/tex]
[tex]9 = \frac{1}{5}x[/tex]
[tex]5 * 9 = 5 * \frac{1}{5}x[/tex]
[tex]45 = x[/tex]

Therefore, Jonas needs 45 minutes to get 90 on his next test.

Their total entertainment expenditures for the month is $42.2 more than the amount budgeted.

The difference between total entertainment expenditures for the month and the amount budgeted is calculated as follows;

Amount budgeted

movies/theater = $20

sporting event = $65

recreation = $22

dining out = $140

Total = $247

Amount spent

movies/theater = $35

sporting event = $32

recreation = $63.8

dining out = $158.4

Total = $289.2

Difference = $289.2 - $247 = $42.2

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Therefore, 698 of the 2510 students have not taken a course in any of these three programming languages using principle of inclusion-exclusion.

We can solve this problem using the principle of inclusion-exclusion. First, we add up the number of students who have taken at least one course:

n(J or L or C) = n(J) + n(L) + n(C) - n(J and L) - n(L and C) - n(J and C) + n(J and L and C)

n(J or L or C) = 1876 + 999 + 345 - 876 - 231 - 290 + 189

n(J or L or C) = 1812

So there are 1812 students who have taken at least one course. Therefore, the number of students who have not taken any of these courses is:

n(not J and not L and not C) = 2510 - n(J or L or C)

n(not J and not L and not C) = 2510 - 1812

n(not J and not L and not C) = 698

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The overall probability is (3/4)^4 * 1/4 = 81/1024, or approximately 0.079.

The probability that Nancy will randomly guess the correct answer for any one of the 5 questions is 1/4 (since there are 4 choices for each question). Since Nancy has not studied at all, her guesses for each question are independent events.

Therefore, the probability that the first question she gets right is question number 5 is the probability that she guesses incorrectly for the first 4 questions (which has a probability of (3/4)^4) and then guesses correctly for question number 5 (which has a probability of 1/4).

Therefore, the overall probability is (3/4)^4 * 1/4 = 81/1024, or approximately 0.079.

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Once the activity is completed, facilitate a discussion among the students to analyze the differences between their predictions and the actual results. This will help reinforce the concept of probability and its properties in their minds.

The teacher should take the following approach with the students who opted to skip the prediction step:

1. Remind the students of the importance of making predictions in the context of probability and its properties. Predictions help them understand the expected outcomes and compare them with the actual results.

2. Instruct the students to pause their coin flipping and return to the prediction step. Ask them to predict the outcomes of the 10 flips, considering the possible results (HH, HT, TH, or TT).

3. After the students have made their predictions, encourage them to resume flipping the coins and recording their findings in a frequency chart.

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The answer is (B): One of the vectors is in the span of the other vectors, and in this case it is vector w that is in the span of vectors u and v.

To determine if one of the given vectors is in the span of the other vectors, we need to check if the vectors are linearly dependent. If they are, then we can express one of the vectors as a linear combination of the others, and that vector is in the span of the others. If they are not linearly dependent, then none of the vectors are in the span of the others.

To check if the vectors are linearly dependent, we can put them into a matrix and row reduce:

[tex]\left[\begin{array}{ccc}1 & 7 & 8 \\-1 & 3 & -5 \\4 & 8 & 6\end{array}\right] \rightarrow\left[\begin{array}{ccc}1 & 7 & 8 \\0 & 10 & 3 \\0 & 0 & -26\end{array}\right][/tex]

We see that the third row is a scalar multiple of the second row, so the vectors are linearly dependent. Therefore, we can express one of the vectors as a linear combination of the others.

Since the third row is a scalar multiple of the second row, we can express the third vector as:

[tex]w--\frac{26}{10} v--\frac{13}{5}\left[\begin{array}{c}-1 \\3 \\-5\end{array}\right][/tex]

So we can express vector w as a linear combination of u and v, and therefore w is in the span of u and v.

Therefore, the answer is (B): One of the vectors is in the span of the other vectors, and in this case it is vector w that is in the span of vectors u and v.

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The area of the figure attached is 15 square meters

The figure is a parallelogram and area of a parallelogram is solved using the formula

= base x height

Information given in the problem includes:

base = 5 m

height = 3 m

plugging in the values in to the formula for area for area of a parallelogram will result to

= 5 m x 3 m

= 15 square meters

Hence the area of the parallelogram  attached is 15 square meters

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In a polynomial function of the lowest degree with rational coefficients that have the given numbers as some of its zeros, we can use the fact that if a number "a" is a zero of a polynomial function, then (x-a) is a factor of the function.

Therefore, we can start by multiplying together (x-a) for each given zero, and then simplify the resulting expression to get the polynomial function in standard form with rational coefficients.

For example, if the given zeros are 2, -1, and 1/2, we would start by multiplying (x-2)(x+1)(x-1/2), which gives:

(x-2)(x+1)(2x-1)

Expanding this expression, we get:

2x^3 - 3x^2 - 5x + 2

This is a polynomial function of degree 3 (highest exponent is 3) with rational coefficients (all coefficients are integers or fractions). Note that this is the lowest degree possible for a polynomial with these zeros, since there are three distinct zeros.

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A graph of each of the function is shown below.

The domain is x = -2, x = 0, x = 1, x = 2.

The range is [-1, -1] U [3, 3].

In Mathematics, a piecewise-defined function is a type of function that is defined by two (2) or more mathematical expressions over a specific domain.

Generally speaking, the domain of any piecewise-defined function simply refers to the union of all of its sub-domains. By critically observing the graph of the given piecewise-defined function, we can reasonably infer and logically deduce that it is constant over the interval x < -2 and x ≥ -2.

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well, is a kite so it has two congruent triangles above and two congruent triangles below, for the perimeter is simply the hypotenuse of all four right-triangles.

[tex]\begin{array}{llll} \textit{using the pythagorean theorem} \\\\ c^2=a^2+o^2\implies c=\sqrt{a^2 + o^2} \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{above}\\ a=\stackrel{adjacent}{4}\\ o=\stackrel{opposite}{3} \end{cases} \\\\\\ above=\sqrt{ 4^2 + 3^2}\implies above=\sqrt{ 16 + 9 } \implies above=\sqrt{ 25 }\implies above=5 \\\\[-0.35em] ~\dotfill[/tex]

[tex]\begin{array}{llll} \textit{using the pythagorean theorem} \\\\ c^2=a^2+o^2\implies c=\sqrt{a^2 + o^2} \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{below}\\ a=\stackrel{adjacent}{3}\\ o=\stackrel{opposite}{7} \end{cases} \\\\\\ below=\sqrt{ 3^2 + 7^2}\implies below=\sqrt{ 9 + 49 } \implies below=\sqrt{ 58 } \\\\[-0.35em] ~\dotfill[/tex]

[tex]5~~ + ~~5~~ + ~~\sqrt{58}~~ + ~~\sqrt{58} ~~ \approx ~~ \text{\LARGE 25.2}[/tex]

AnswerAnswer:1/11

Step-by-step explanation:

The researcher is conducting a descriptive statistical analysis using a frequency distribution. In this analysis, the frequency refers to the number of times each response option (categories) on the Likert-type scale is selected by the participants.

The categories represent the different response options on the Likert scale, which typically range from strongly agree to strongly disagree or similar variations. By analyzing the frequency distribution of responses across these categories, the researcher can identify patterns, trends, and central tendencies in the data, which can be useful for understanding the overall attitudes or opinions of the participants regarding the topic being investigated. '

This type of analysis is often presented in the form of a table or bar chart, displaying the frequency of each response category, making it easy to interpret and visualize the results.

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The requried distance between points P(1,6) and Q(5,8) in simplest radical form is 2√5.

We can use the distance formula to find the distance between the two points:

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

where (x₁, y₁) = (1, 6) and (x₂, y₂) = (5, 8).

Substituting the values, we get:

d = √[(5 - 1)² + (8 - 6)²]

= √[4² + 2²]

= √(16 + 4)

= √20

= 2√5

Therefore, the distance between points P(1,6) and Q(5,8) in simplest radical form is 2√5.

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False. When performing regression analysis, different samples will not necessarily yield the same sample regression line.

This is because the sample regression line is based on the specific data points in the sample, and different samples may have different data points that can lead to different regression lines. However, if the samples are representative of the same population and the relationship between the variables is consistent, the regression lines may be similar. The term "content loaded" does not apply to this question.

False. When performing regression analysis, different samples will not all yield the same sample regression line. The regression line depends on the specific data points in the sample, so if you have different samples, it's likely that their regression lines will be different as well.

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To calculate the reorder point, we need to consider the lead time (6 days) and the average daily usage of the product. Assuming a 30-day month, the average daily usage can be calculated as:
Product used per day = Total product used in a month / Number of days in a month
Product used per day = Inventory / 30

To determine the reorder point, we need to know the daily consumption rate of the product and the lead time (number of days it takes for the order to arrive).

1. Calculate the daily consumption rate: Since we have a 30-day month, we will divide the monthly consumption by 30.
Total consumption in a month = X gallons (Information not provided)
Daily consumption rate = X gallons / 30 days

2. Calculate the lead time demand: Since it takes 6 days for the order to arrive, we need to find out how many gallons are consumed during this time.
Lead time demand = Daily consumption rate * Lead time
Lead time demand = (X gallons / 30 days) * 6 days

3. Determine the reorder point: The reorder point is the level of inventory at which the product should be reordered to ensure that there is enough stock to cover the lead time demand.
Reorder point = Lead time demand
Reorder point = (X gallons / 30 days) * 6 days

Let's assume that the reorder point is the level of inventory at which we need to place a new order to avoid running out of stock. So, the reorder point can be calculated as:

Reorder point = Lead time demand + Safety stock

Here, the lead time demand is the product used during the lead time (6 days) and the safety stock is the buffer inventory kept to avoid stockouts. Let's assume a safety stock of 50 gallons.

Lead time demand = Product used per day * Lead time
Lead time demand = (Inventory / 30) * 6
Lead time demand = Inventory / 5

Reorder point = (Inventory / 5) + 50

We need to solve for the inventory level at which we should reorder. Let's assume the answer is X.

X / 5 + 50 = X
X - X/5 = 50
4X/5 = 50
X = (5/4) * 50
X = 62.5

So, the reorder point is 62.5 gallon. Therefore, the correct answer is "none of the above".

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To solve this problem, we can set up an integral to find the area of the region s. Let's assume that the line passing through the origin divides the region into two equal parts. We can express this line as y = mx, where m is the slope of the line.

To find the equation that gives the value of m, we can set up an integral that represents the area of one of the two regions. Let's call the area of one of the regions A. Since the line passes through the origin, the boundaries of the integral for this region are 0 and some value of x, which we'll call x0.

The equation for the line is y = mx, so the equation for the curve that defines the upper boundary of this region is y = mx. We can solve for x in terms of y to find the limits of integration for this region:

x = y/m

Since we want to find the area of this region, we can integrate with respect to y:

A = ∫0x0 (x/y) dy

Now we need to set up a similar integral to find the area of the other region. Let's call the area of the other region B. Since the two regions have equal areas, we know that A = B.

The boundaries of the integral for the second region are x0 and some value of x, which we'll call x1. The equation for the line is still y = mx, so the equation for the curve that defines the upper boundary of this region is y = mx. We can solve for x in terms of y again:

x = y/m

And integrate with respect to y:

B = ∫x0x1 (x/y) dy

Now we can set up an equation that equates the two regions:

A = B

∫0x0 (x/y) dy = ∫x0x1 (x/y) dy

We can solve this equation for m:

m = x1/(2x0)

So the equation involving one or more integrals whose solution gives the value of m is:

m = x1/(2x0)

where x0 and x1 are the limits of integration for the two regions. We could solve this equation by finding the value of x1 that satisfies the condition that the two regions have equal areas, given a value of x0. However, we are only asked to write the equation, not solve it.

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