What is the answer to the question?

Answer: 94 square inches 2

Step-by-step explanation:

2×(5×4 + 5×3 + 4×3) = 94 sq in 2

length x width x height

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Minimum value is 50, maximum value is 66, median is 59, Q1 is 53.5 and Q3 is 65. Box-and-whisker plot:

A box plot, often referred to as a whisker plot, shows the distribution of a continuous variable along with details like the median, quartiles, and outliers. A set of data's variety is shown graphically in this way. The plot, often known as a five-number summary, presents five bits of information. The fundamental benefit of box whisker plots is that the overlap between the distributions of various qualities may be calculated by comparing them side by side.

You must determine the minimum value, maximum value, median, and quartiles in order to create a box-and-whisker plot of the provided data.

The base number is fifty.

The highest number is 66.

You must sort the data from smallest to largest in order to determine the median. The listed data are:

50, 51, 51, 52, 53, 53, 54, 55, 58, 58, 58, 60, 61, 62, 63, 65, 65, 66, 66

The middle value of the ordered data is the median. Finding the average of the two middle values is necessary since the data set (18 values) contains an even number of values:

(58 + 60) ÷ 2 = 59

Divide the ordered data into four equal sections before calculating the quartiles. The median of the lower half of the data set is the first quartile (Q1). The median of the data set's upper half represents the third quartile (Q3).

The data set's lower half consists of:

50, 51, 51, 52, 53, 53, 54, 55

The data set's upper half is as follows:

58, 58, 58, **60**, **61**, **62**, **63**, **65**, **65**, **66**, **66**

First quartile (Q1):

(53 + 54) ÷ 2 = **53.5**

Third quartile (Q3) comprises:

(65 + 65) ÷ 2 = **65**

With all of this knowledge at hand, you can now develop a box-and-whisker plan.

Given below:

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The two statements that completes the above proof are:

ΔABC ~ ΔDEC - Side - Angle - Side Postulate

∠1≅∠2 - Corresponding angles.

The Side-Angle-Side (SAS) postulate states that if two sides and the angle between them in one triangle are congruent to two sides and the angle between them in another triangle, then the triangles are congruent.

Corresponding angles are equal because of the congruence of the triangles. Two angles are corresponding if they occupy the same relative position in two distinct intersecting lines, such that they are on the same side of the transversal and are in the same position in relation to the intersection. Corresponding angles are equal if the two intersecting lines are parallel.

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Answer:

5x

6xy

2x3y

hope it's helpful

Estimate the mean number of travel days per year for salespeople employed by three hardware distributors with a 0.90 degree of confidence and an error margin of 2 days, a sample size of at least 179 salespeople should be selected.

The sample size needed to estimate the population mean with a 0.90 degree of confidence:

[tex]n = (Z\times\sigma / E)^2[/tex]

where:

Z = the Z-score for the desired level of confidence (0.90 corresponds to a Z-score of 1.645)

σ = the population standard deviation (unknown)

E = the maximum error of the estimate, which is given as 2 days

We can estimate the population standard deviation using the sample standard deviation from the pilot study, which is given as 16 days.

Therefore, plugging in the values, we get:

[tex]n = (1.645 \times 16 / 2)^2 = 178.26[/tex]

Rounding up to the nearest whole number, we get a minimum sample size of 179 salespeople.

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The measure of the angle m∠FGH is 83°.

Since the measure of an angle formed by a two tangents drawn from a point outside the circle is one-half the the difference of the intercepted arcs.

Thus, we can say:

m∠FGH = 1/2 * (∠FIH - 97°)

∠FIH = 360 - 97 = 263° (sum of angle in a circle)

m∠FGH = 1/2 * (263 - 97°)

m∠FGH =  1/2 * 166

m∠FGH = 83°

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The measures of center and variability that would be most appropriate to describe the given distribution is D) Median and IQR.

The middle value in a set of data is represented by the median, a measure of central tendency. It is the value that, when a dataset's values are ranked from lowest to highest, distinguishes the lower from the upper half of the dataset.

The middle 50% of a dataset's interquartile range (IQR) is a measure of variability that captures this dispersion. It is the difference between the data's first (Q1) and third (Q3) quartiles. The number that divides the lowest 25% of the data from the remaining data is known as the first quartile, and the value that separates the highest 25% of the data from the remaining data is known as the third quartile.

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The complete question is:

If we consider grade "C" or above as a pass grade, then the number of  students who successfully passed the course are 10 students.

The Number of students who got grade "A" = 8,

The Number of students who got "B" grade is 2,

The Number of students who got grade "C" = 0,

The Number of students who got grade "D" = 0,

The Number of students who got grade "F" = 1,

Since the question asks for the number of students who successfully passed the course (i.e., grade "C" or above is considered a pass grade),

We add the number of students who got grades "A" and "B".

So, total number of students who successfully passed course are :

⇒ 8 (students who got grade "A") + 2 (students who got grade "B")

⇒ 10

Therefore, 10 students successfully passed the course.

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The given question is incomplete, the complete question is

If 8 students of the class got grade "A", 2 student got grade "B", No student got grade "C" and "D", 1 student got grade "F".

Considering grade "C" or above as a pass grade, how many students from this data successfully passed the course?

For all given Quadrilaterals none has enough information provided in the problem to definitively say - if they are a parallelogram.Hence, option E is correct for all.

If a quadrilateral is a parallelogram, it satisfies the following properties:

It is important to note that in order to definitively conclude that a quadrilateral is a parallelogram, all four properties must be satisfied. If only one or some of the properties are met, it does not necessarily guarantee that the quadrilateral is a parallelogram.

Figure 1-: Adjacent angles are 105 and 75 degrees, which satify the condition of opposite angles being equal but except this no other information is provided, Hence, we don't have enough information to say figure 1 is a parallelogram.

Figure 2-: Diagonals bisect each other and make angle 65 degree with each other. Given that Diagonal 1 bisects Diagonal 2 and the opposite sides of the quadrilateral are equal, by using SAS criterion, the congruency of triangles formed by the diagonals can be derived to say that the opposite angles of the quadrilateral are also equal . However, we still need more information about parallelism of sides to definitively say given quadrilateral is a parallelogram.

Figure 3-: One pair of opposite side is equal in length ,while other pair of line is parallel to each other.This information is insufficient to determine if given quadrilateral is a parallelogram.

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Answer:

$21.28

Step-by-step explanation:

We Know

Dylan bought 3 identical shirts online for a total cost of $71.83

$7.99 for shipping

We have the equation:

71.83 = 3s + 7.99

63.84 = 3s

s = $21.28

So, each shirt cost $21.28

The area swept clean by the rear window wiper blade is approximately 91.37 square inches.

We'll calculate the area swept clean by the rear window wiper blade using the given information.
Convert the angle from degrees to radians:

115 degrees * (π radians / 180 degrees) = (23π/36) radians.
Find the length of the arc that the wiper blade traces out:

Arc length = (radius * angle) = (10 inches * (23π/36) radians) ≈ 20.08 inches.
Calculate the area of the sector formed by the arc and the pivot point:

Sector area = (1/2) * (radius² * angle) = (1/2) * (10 inches² * (23π/36) radians) ≈ 100.42 square inches.
Find the length of the arc that the mounting point of the blade traces out:

Arc length = (radius * angle) = (3 inches * (23π/36) radians) ≈ 6.02 inches.
Calculate the area of the sector formed by this arc and the pivot point:

Sector area = (1/2) * (radius² * angle) = (1/2) * (3 inches² * (23π/36) radians) ≈ 9.05 square inches.
Subtract the smaller sector area from the larger one to get the area swept clean by the wiper blade:

Area swept clean = 100.42 square inches - 9.05 square inches ≈ 91.37 square inches.

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The area swept clean by the rear window wiper blade is approximately 31.53 square inches.

To determine the area swept clean by the rear window wiper blade, we first need to find the length of the arc covered by the blade. We can use the formula:

arc length = radius * angle in radians

To convert the angle from degrees to radians, we multiply by[tex]\pi /180[/tex]:

115 degrees x [tex]\pi /180[/tex] = 2.007 radians

The radius is the length of the wiper arm, which is 13 inches minus the distance from the pivot point to the blade, which is 3 inches. Therefore, the radius is:

13 inches - 3 inches = 10 inches

Now we can calculate the arc length:

arc length = 10 inches x 2.007 radians = 20.07 inches

So the area swept clean by the wiper blade is approximately equal to the area of the sector formed by the arc and the two radii. The formula for the area of a sector is:

area = (angle in radians / [tex]2\pi[/tex]) x [tex]\pi r^2[/tex]

Plugging in values we have:

area = (2.007 / [tex]2\pi[/tex]) x [tex]\pi[/tex][tex](10 inches)^2[/tex] = 31.53 square inches

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"It can vary - it really depends on the distribution of the variable."

The number of intervals, or bins, to choose when creating a histogram can vary depending on the distribution of the variable.

Most often, about 8-10 intervals are used, but there is no set rule. It is generally recommended to have at least 5 intervals, but if the data is highly skewed or has outliers, more intervals may be needed to accurately represent the distribution.

Ultimately, the goal is to choose a number of intervals that provides a clear visual representation of the data without oversimplifying or overcomplicating the histogram.

The number of intervals or bins to be chosen when creating a histogram can vary and it really depends on the distribution of the variable.

While most often, about 8-10 bins are used, there is no hard and fast rule for the number of bins to be used in a histogram.

In general, the number of bins should be large enough to display the shape of the distribution clearly, but not so large that it obscures important features of the distribution or leads to overfitting.

A minimum of 5 bins is recommended to display the basic shape of the distribution, but more bins may be necessary for complex or multi-modal distributions.

Depending on the distribution of the variable, a histogram's number of intervals or bins can be altered.

There is no established guideline, however 8–10 intervals are typically utilized.

A minimum of five intervals are often advised, however if the data is extremely skewed or contains outliers, more intervals could be required to correctly depict the distribution.

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The required sample size of the given estimate is 1090.9809

The population standard deviation is approximately σ = 50.2. You would like to be 90% confident that your estimate is within 1.5 of the true population mean.

Statistics is the study of mathematics that deals with relations between comprehensive data.

Here,

For the 90% confidence the value of the z = 1.645

Now the sample size is given as,

n = [z × σ/2.5]²

Substitute values in the above equation,

n = [1.645×50.2/2.5]²

n = 1090.9809

Thus, the required sample size of the given estimate is 1090.9809.

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Answer: A. 5,220,000 songs were downloaded in the year 2020.

Step-by-step explanation: The interpretation of s(2020) = 5,220,000 is that in the year 2020, 5,220,000 songs were downloaded. The function s(d) represents the number of songs downloaded in a year d, so plugging in 2020 for d gives us the specific number of songs downloaded in that year.

The answer is A. Evaluation researchers encounter more logistical problems than other researchers because evaluation research occurs in the context of real life.

Evaluation research often takes place in real-world settings, which can present logistical challenges such as accessing participants, coordinating schedules, and dealing with unexpected events. Additionally, evaluation research often involves multiple stakeholders and requires collaboration and communication among various groups, which can further complicate logistical issues. While evaluation research may also involve longer timelines, higher costs, measurement problems, and examination of multiple variables, these factors do not necessarily contribute to greater logistical challenges.

This means that evaluation researchers have to navigate complex, real-world situations, adapt to unforeseen challenges, and work with various stakeholders, making the research process more logistically challenging compared to controlled laboratory settings or theoretical research.

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Evaluation research is a type of research that focuses on assessing the effectiveness, efficiency, and impact of programs, policies, or interventions in real-life settings.

The correct answer is A. occurs in the context of real life.

This often involves evaluating the outcomes and impacts of interventions in complex and dynamic environments, such as organizations, communities, or systems. As a result, evaluation researchers may encounter more logistical problems compared to other types of researchers because they need to navigate real-life contexts, deal with multiple stakeholders, collect data from diverse sources, and address issues such as ethics, confidentiality, and validity in the evaluation process. Logistical problems may include challenges related to data collection, measurement, sample selection, data quality, and managing time and resources.

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Answer:

1.504 x 10^4

Step-by-step explanation:

The correct statement about the graph is,

A graph of two solid lines that intersect at point (4, 0).

Given that;

This system of inequalities models the scenario.

4x + 2y ≤ 16

And, x + y ≥ 4

Now, We can draw the inequalities in graph, we get;

The intersection point of the inequality is, (4, 0)

Thus, The correct statement about the graph is,

A graph of two solid lines that intersect at point (4, 0).

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The probability that a randomly selected can in the box is dented or contains vegetables is 19/20.

To solve the problem, we need to use the formula for the probability of the union of two events:

P(A or B) = P(A) + P(B) - P(A and B)

where A and B are two events.

In this case, we want to find the probability that a randomly selected can is dented or contains vegetables. Let's call these events D and V, respectively.

We are given that there are 20 cans in total, 12 of which contain vegetables, and 5 of which are dented, with 2 of the dented cans containing vegetables. This information allows us to calculate the probabilities of the individual events

P(D) = 5/20 = 1/4

P(V) = 12/20 = 3/5

P(D and V) = 2/20 = 1/10

To find the probability of the union of these events, we plug these values into the formula

P(D or V) = P(D) + P(V) - P(D and V)

= 1/4 + 3/5 - 1/10

= 19/20

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The correct answer is (c) 1.9

To calculate the test statistic in this example, we can use the formula for a z-test for proportions:

[tex]z = (p - P) / \sqrt{[P(1-P) / n][/tex]

where:

p is the sample proportion of vegetarians (140/1000 = 0.14)

P is the hypothesized population proportion of vegetarians (0.12)

n is the sample size (1000)

Substituting these values into the formula, we get:

[tex]z = (0.14 - 0.12) / \sqrt{[0.12(1-0.12) / 1000][/tex]

z = 0.02 / 0.0110

z = 1.818 (rounded to three decimal places)

Therefore, the test statistic in this example will be approximately equal to 1.818. The answer closest to this value is (c) 1.9

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Point A = (5,4). If you rotated A 90 degrees about the point (2,-1), the coordinates of A' is (-1,2).

Rotation is the process of rotating an object or a point around a fixed point or axis. In mathematics, rotation refers to a transformation that preserves the size and shape of an object while changing its orientation. It is a basic geometric transformation that is used in various fields, including mathematics, physics, engineering, and computer graphics.

In a two-dimensional space, a rotation is typically described by an angle of rotation and a fixed point, which is known as the center of rotation. The angle of rotation represents the amount by which the object is rotated, while the center of rotation is the point around which the object is rotated. In a three-dimensional space, a rotation is described by an axis of rotation and an angle of rotation.

To rotate point A 90 degrees counterclockwise about the point (2,-1), we can use the following formula:

A' = (x', y') = (a + (x - a) cosθ - (y - b) sinθ, b + (x - a) sinθ + (y - b) cosθ)

where (a,b) is the center of rotation and θ is the angle of rotation (90 degrees in this case).

Substituting the given values, we get:

a = 2, b = -1, x = 5, y = 4, θ = 90 degrees

x' = 2 + (5 - 2) cos(90) - (4 + 1) sin(90) = 2 - 3 = -1

y' = -1 + (5 - 2) sin(90) + (4 + 1) cos(90) = -1 + 3 = 2

Therefore, the coordinates of A' are (-1, 2).

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The output of the LTI system with frequency response function [tex]h(w) = 1/(jw^3)[/tex], given an input[tex]x(t)[/tex], is:

[tex]y(t)[/tex]= inverse Fourier transform of [tex][1/(jw^3) X(w)][/tex]

To compute the output of an LTI system with frequency response function [tex]h(w) = 1/(jw^3)[/tex], given an input x(t), we can use the Fourier transform:

[tex]H(w)[/tex] = Fourier transform of [tex]h(t)[/tex]

[tex]X(w)[/tex] = Fourier transform of [tex]x(t)[/tex]

[tex]Y(w) = H(w) X(w)[/tex]

[tex]Y(w)[/tex] is the Fourier transform of the output [tex]y(t)[/tex].

Using the given frequency response function, we have:

[tex]H(w) = 1/(jw^3)[/tex]

Taking the Fourier transform of the input [tex]x(t)[/tex], we have:

[tex]X(w)[/tex] = Fourier transform of [tex]x(t)[/tex]

And multiplying [tex]H(w)[/tex] and[tex]X(w)[/tex], we get:

[tex]Y(w) = H(w) X(w) = 1/(jw^3) X(w)[/tex]

Taking the inverse Fourier transform of [tex]Y(w)[/tex], we get the output [tex]y(t)[/tex]:

[tex]y(t)[/tex] = inverse Fourier transform of [tex]Y(w)[/tex]

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The critical value used to test if there is evidence of an effect due to blocks at the 5% level of significance is 10.76.

The critical value can be determined using a statistical table.

The degrees of freedom for the blocks is 4 (df_b = 5 - 1 = 4) and for the treatments is 2 (df_t = 3 - 1 = 2).

The critical value for a 5% level of significance is 10.76.

This value can be found in the statistical table given in the textbook.

The critical value used to test if there is evidence of an effect due to blocks at the 5% level of significance can be calculated by using the F-test statistic.

The F-test is used to compare the variance between the blocks (SSBL) and the variance between the groups (SSB).

The F statistic is calculated by dividing the variance between the blocks (SSBL) by the variance between the groups (SSB).

In this case,

The F statistic is 3738/1048.93 = 3.56.

The critical value for a 5% level of significance can be found using a statistical table.

According to the table,

The critical value for an F statistic of 3.56 with four degrees of freedom in the numerator and four degrees of freedom in the denominator is 10.76

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The value of a for which the solution of the inequality is all real numbers is a = 5.

We have to first simplify the inequality:

A(x + 3) < 5x + 15 - x

Ax + 3A < 4x + 15

Ax - 4x < 15 - 3A

Simplifying further:

x(A - 4) < 15 - 3A

Now, we need the inequality to hold for all real values of x. This means that the coefficient of x, (A - 4), must have a fixed sign, and that the right-hand side, 15 - 3A, must be unbounded.

For the coefficient of x to have a fixed sign, we need either A - 4 < 0 or A - 4 > 0. This means that A must be less than 4 or greater than 4.

For the right-hand side to be unbounded, we need 15 - 3A to be equal to positive or negative infinity. Since 15 - 3A is a linear function of A, this only occurs when A = 5.

So, the solution to the inequality for all real numbers is:

if A < 4, then the solution is x < (15 - 3A) / (4 - A)

if A > 5, then the solution is x > (15 - 3A) / (4 - A)

if A = 5, then the solution is all real numbers.

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The height of the pyramid whose volume is already given would be = 1,440,000 ft

To calculate the height of the pyramid, the formula of the Volume of the pyramid should be used.

Volume = 1/3lwh

where l = base length = 8h-30 ft= 3.75 ft

w = base width= 8h+20 ft = -2.5 ft

h = h ft

The volume of the pyramid = 4,500,000 cubic feet

4,500,000 = 1/3 × (3.75)(-2.5)h

Make h the subject of formula;

h = 4500000×3/3.75×(-2.5)

h = 13500000/9.375

h = 1,440,000 ft.

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The sample mean for 100 pell grant recipients is more likely to be within $80 of $2,600 for State A, given the relatively smaller population size and the use of the t-distribution.

To compare the two states, we need to calculate the probability of the sample mean for 100 pell grant recipients being within $80 of $2,600 for each state.

For State A, with only 4,000 pell grant recipients, the sample size of 100 is relatively large, but not enough to use the normal distribution. Therefore, we need to use the t-distribution, which has fatter tails than the normal distribution, making it more likely to produce extreme values. This means that the probability of the sample mean for 100 pell grant recipients being within $80 of $2,600 is higher for State A than for State B.

For State B, with a larger population of 200,000 pell grant recipients, the sample size of 100 is a small fraction of the population, allowing us to use the normal distribution. However, as the distribution is narrower than the t-distribution, the probability of the sample mean for 100 pell grant recipients being within $80 of $2,600 is lower for State B than for State A.

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Answer:

-6/7

Step-by-step explanation:

Answer:

$481

Step-by-step explanation:

37×$17=$481

he made $481

[tex]\textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=2827.43\\ h=4 \end{cases}\implies 2827.43=\pi r^2(4)\implies \cfrac{2827.43}{4\pi }=r^2 \\\\\\ \sqrt{\cfrac{2827.43}{4\pi }}=r\hspace{5em}\stackrel{\textit{twice that is the \underline{diameter}}}{2\sqrt{\cfrac{2827.43}{4\pi }}} ~~ \approx ~~ \text{\LARGE 30}[/tex]