angles homework please help

Answer:

some numbers that involve the

a=3

b=4

c=5

The box plot of the assembly line: 5.2,5.4,5.3,5.3,5.4,5.6,5.7,5.6,5.7 are;

A box plot is a graphical representation of the distribution of a dataset through their quartiles. The approximation method is a way to estimate the exact values, and it's usually represented through a diagram. The data given is the diameter of cans in an assembly line: 5.2, 5.4, 5.3, 5.3, 5.4, 5.6, 5.7, 5.6, 5.7

To construct a box plot, you need to follow some steps:Arrange the data in order from smallest to largest.Find the median of the dataset. This will be the line that divides the box into two parts.

Find the lower quartile (Q1) by splitting the dataset into two equal parts and finding the median of the lower half.

Find the upper quartile (Q3) by splitting the dataset into two equal parts and finding the median of the upper half.

Calculate the interquartile range (IQR) which is the difference between the upper and lower quartiles (IQR = Q3 - Q1).Find the lower limit of the dataset, which is the smallest number in the set greater than or equal to Q1 - 1.5 × IQR

.Find the upper limit of the dataset, which is the largest number in the set less than or equal to Q3 + 1.5 × IQR

.Draw a vertical line from the lower limit to the lower whisker, which is the smallest value in the dataset that is greater than or equal to the lower limit

.Draw a vertical line from the upper limit to the upper whisker, which is the largest value in the dataset that is less than or equal to the upper limit.

Draw a box with the bottom at Q1, the top at Q3, and the line for the median in the middle. Label the lower whisker, the upper whisker, the median, Q1, and Q3. The box plot of the given data is:

The lower whisker is at 5.2 and the upper whisker is at 5.7. The median is at 5.4. Q1 is at 5.3, and Q3 is at 5.6.

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Al-Khwarizmi made significant contributions to the world of mathematics, particularly in the field of algebra. His major accomplishment was the development of algebra as an independent mathematical discipline.

Al-Khwarizmi, a Persian mathematician and scholar, wrote the book titled "Kitab al-Jabr wal-Muqabala," which laid the foundation for modern algebra. In this influential work, he introduced systematic methods for solving linear and quadratic equations. Al-Khwarizmi's algebraic techniques, including the use of variables and equations, greatly advanced mathematical understanding and problem-solving. His work also contributed to the development of algorithms and mathematical notation, such as the concept of the "zero" and the decimal system. Overall, al-Khwarizmi's contributions revolutionized mathematics and had a profound impact on subsequent generations of mathematicians and scientists.

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Answer:

if youre talking about line OA, it is

x - 2y =0

Answer:

-1/2

Step-by-step explanation:

Use the formula m=y2-y1/x2-x1

m= -4 -1/3- -7

m= -4+-1/ 3++7

m=-5/10

m=-1/2

(a) The proportion of students who scored at least 26 points on the test is 0.0228

(b) The 31st percentile of the distribution of test scores is 20.12

(a) To find the proportion of students who scored at least 26 points on the test, we need to find the area under the normal distribution curve to the right of 26.

Using a standard normal table, we can find the corresponding z-score

z = (26 - 22) / 2 = 2

The area to the right of z = 2 is approximately 0.0228.

Therefore, the proportion of students who scored at least 26 points on the test is 0.0228 (rounded to five decimal places).

(b) To find the 31st percentile of the distribution of test scores, we need to find the z-score that corresponds to the 31st percentile.

Using a standard normal table, we can find the z-score that corresponds to a cumulative area of 0.31, which is approximately -0.44.

Then, we can solve for the corresponding test score

-0.44 = (x - 22) / 2

x = 20.12

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The given question is incomplete, the complete question is:

The distribution of scores on a recent test closely followed a Normal Distribution with a mean of 22 points and a standard deviation of 2 points. For this question, Do not apply the standard deviation rule.

(a) What proportion of the students scored at least 26 points on this test, rounded to five decimal places?

b)What is the 31 percentile of the distribution of test scores, rounded to three decimal places?

c. The mean of 16.91 and the standard deviation of 0.12 are parameters.

In statistics, parameters are values that describe a population. In this case, the mean ounces and the standard deviation calculated from the sample are used to estimate the corresponding parameters of the population. Since the sample was taken from a larger population of bottles of water, the mean of 16.91 ounces and the standard deviation of 0.12 ounces are estimates of the true population parameters. Therefore, they are considered parameters rather than statistics.

A statistic, on the other hand, is a value calculated from a sample and is used to describe or estimate a population parameter. In this scenario, the values calculated from the sample (mean and standard deviation) are used as estimates of the population parameters, making them parameters rather than statistics.

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The meaning of vertical intercept in this context can be inferred from the options to mean that the helicopter took off from a height of 50 ft.

In a linear problem or a graph, the vertical intercept, which is also known as the y-intercept, refers to the value of y when the value of x is 0.

Assuming y in this problem refers to the height and x refers to the minutes, the vertical intercept would be the height at 0 minutes.

If the vertical intercept is 50 ft then it means that the helicopter took off from 50 feet because at takeoff, the value of x would be 0 minutes.

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The area of the given triangle is 87.55 square centimeter.

From the given triangle, base=17 cm and height=10.3 cm.

The basic formula for the area of a triangle is equal to half the product of its base and height, i.e., A = 1/2 × b × h.

Here, area of a triangle = 1/2 ×17×10.3

= 87.55 square centimeter

Therefore, area of the given triangle is 87.55 square centimeter.

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The alternative and hull hypotheses is H₀: P=0.03, H₁: P≠0.03.

Given that,

According to a representative of the Ohio Department of Education, 3% of graduating high school seniors in Ohio have dropped out in recent years. Out of the 600 pupils enrolled, 30 students from Podunk High School dropped out of school last year.

We have to find is there enough data to say that this school's dropout rate is higher than the state average. Using symbols and words, state the alternative and hull hypotheses.

We know that,

n is 600,

3% Ohio high school seniors dropout.

30 dropout out of 600 in Podunk high school.

Here,

H₀: μ₁=μ₂  vs   H₁:μ₁≠μ₂

In coordinates,

H₀⇒ dropout number is equal for both schools.

H₁⇒ dropout number not equal for both schools.

H₀: P=0.03, H₁: P≠0.03.

Therefore, The alternative and hull hypotheses is H₀: P=0.03, H₁: P≠0.03.

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The statement m∠WXY + m∠YXZ = 180° is true.

From the diagram, m∠WXY > m∠YXZ ⇒ m∠WXY ≠ m∠YXZ. So, option (a) is not true.

From the diagram, m∠WXY > 90° and m∠YZX < 90° ⇒ m∠WXY ≠ m∠YXZ. So, option (b) is not true.

The line WXZ is a straight line so ∠WXY and ∠YXZ are Linear pair. The sum of collinear angles is 180°. So, option (c) is correct.

From the diagram, m∠YXZ ≠ m∠XYZ but since m∠WXY + m∠YXZ = 180° ⇒ m∠WXY + m∠XYZ ≠ 180°. So, option (d) is not correct.

Option (c) is, thus, the correct option.

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Disclaimer: The question was incomplete. Please find the dull content below.

Which statement regarding the attached diagram is true?

m∠WXY = m∠YXZ

m∠WXY < m∠YZX

m∠WXY + m∠YXZ = 180°

m∠WXY + m∠XYZ = 180°

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We must represent the number 4.082, the options you are giving us are from an apparent decomposition of the previous number

The correct option is B: Four and eighty-two thousandths, It is indeed 4 whole units and 82 thousandths. The other options are not correct

Answer: c = 5n

Step-by-step explanation:

Cost of movie ticket = $5 per person

Number of people attending the movie = n

The total cost c, will be the multiplication of the cost of movie ticket per person by the total.number of people attending the movie. This will be:

c = 5 × n

c = 5n

To evaluate the given integral, we can use the fact that D is symmetric with respect to both axes. This means that we can rewrite the integral as:
∫∫D g(x^2 tan(z) + ã¡ + 3) dA
= 4 ∫∫D g(x^2 tan(z) + ã¡ + 3) dxdy where we have used the symmetry of D to change the limits of integration.

Next, we can convert to polar coordinates, which is particularly convenient since D is defined in terms of x and y in terms of their distance from the origin. Specifically, we can substitute x = r cos(θ) and y = r sin(θ), and also use the identity tan(z) = sin(z) / cos(z) = y/x:
∫∫D g(x^2 tan(z) + ã¡ + 3) dxdy
= 4 ∫θ=0^2π ∫r=2^√(9 - r^2 cos^2(θ)) 0 g(r^2 sin(θ) cos(θ) + ã¡ + 3) rdrdθ where we have used the equation of the circle r^2 = x^2 + y^2 = 4 to set the limits of integration for r.
Now we can simplify the integrand by noting that r^2 sin(θ) cos(θ) = (r^2/2) sin(2θ). We also have the constant term ã¡ + 3, which does not depend on r or θ. Thus, we can pull it out of the integral:
4 ∫θ=0^2π ∫r=2^√(9 - r^2 cos^2(θ)) 0 g((r^2/2) sin(2θ) + ã¡ + 3) rdrdθ
= (ã¡ + 3) 4 ∫θ=0^2π sin(2θ) dθ ∫r=2^√(9 - r^2 cos^2(θ)) 0 g(r^2/2) rdr
The inner integral can be evaluated using the substitution u = r^2/2, so that rdr = du/sqrt(2), and the limits of integration become u = 0 to u = 9/2 cos^2(θ). We can then use the fact that g is an arbitrary function, so we can write:
∫r=2^√(9 - r^2 cos^2(θ)) 0 g(r^2/2) rdr
= ∫u=0^9/2cos^2(θ) g(u) du/sqrt(2)
Finally, we can substitute back into the original expression:
4 ∫θ=0^2π sin(2θ) dθ ∫r=2^√(9 - r^2 cos^2(θ)) 0 g(r^2/2) rdr
= (ã¡ + 3) 4 ∫θ=0^2π sin(2θ) dθ ∫u=0^9/2cos^2(θ) g(u) du/sqrt(2)
= 2π (ã¡ + 3) ∫u=0^9/2 g(u) du

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Coating the seed typically improves seed quality and germination rates, so the same seeding rate can be maintained even with coated seed.

To convert mass-based seeding rates to population-based establishment rates, we need to consider the area conversion factors provided:

a. Given a seeding rate of 60 lbs/acre for cereal rye, we can convert it to seeds per square foot by using the conversion factor of 1 acre = 43,560 square feet.

b. Given a seeding rate of 15 kg/ha for crimson clover, we can convert it to seeds per square meter using the conversion factor of 1 hectare = 10,000 square meters.

To convert population-based seeding rates to mass-based seeding rates, we can use the given area conversion factors:

a. Given a seeding rate of 30 plants/square foot for oats, we can convert it to pounds of seed per acre.

b. Given a seeding rate of 20 plants/square meter for hairy vetch, we can convert it to kilograms of seed per hectare.

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The differential equation is y(x) = C₁ \(x^{-1/3}\) + C₂ \(x^{-1}\) + u₁(x)\(x^{-1/3}\) + u₂(x)x⁻¹.

To solve the given differential equation using the method of variation of parameters, we will follow these steps:

Find the complementary solution by solving the associated homogeneous equation: 3x²y" + 7xy' + y = 0.

Assume a particular solution in the form of \(y_p\) = u₁(x)y_1(x) + u₂(x)y₂(x), where y₁ and y₂ are solutions of the homogeneous equation and u₁(x) and u₂(x) are unknown functions to be determined.

Substitute the particular solution into the original differential equation and solve for u₁'(x) and u₂'(x).

Integrate u₁'(x) and u₂'(x) to find u₁(x) and u₂(x).

Substitute the found values of u₁(x) and u₂(x) back into the particular solution to obtain the general solution of the differential equation.

Let's proceed with the steps:

The associated homogeneous equation is 3x₂y" + 7xy' + y = 0. We can try to find a solution in the form of y = \(x^r\). By differentiating twice and substituting into the equation, we get the characteristic equation:

\(3r(r-1)x^r + 7rx^r + x^r = 0\)

Simplifying, we have:

3r² - 3r + 7r + 1 = 0

3r² + 4r + 1 = 0

Factoring the quadratic equation, we find:

(3r + 1)(r + 1) = 0

This gives us two roots: r = -1/3 and r = -1.

Therefore, the complementary solution is \(y_c = C_1 x^{-1/3} + C_2 x^{-1}\), where C₁ and C₂ are constants.

Now, we assume the particular solution in the form of \(y_p = u_1(x)y_1(x) + u_2(x)y_2(x)\), where y₁(x) and y₂(x) are the solutions of the homogeneous equation we found in step 1.

Since we found two solutions, \(y_1(x) = x^{-1/3}\) and \(y_2(x) = x^{-1}\), we can write the particular solution as:

\(y_p = u_1(x) x^{-1/3} + u_2(x) x^{-1}\)

Substituting \(y_p\) into the original differential equation, we have:

\(3x^2(u_1''(x) x^{-1/3} + u_2''(x) x^{-1}) + 7x(u_1'(x) x^{-1/3} + u_2'(x) x^{-1}) + u_1(x) x^{-1/3} + u_2(x) x^{-1} = x^2\)

Simplifying, we get:

\(3u_1''(x) + 3u_1'(x)x^{-4/3} + 7u_1'(x) + 7u_1(x)x^{-4/3} + 3u_2''(x)x^{-2} + 7u_2'(x)x^{-2} + u_2(x)x^{-2} = x^2\)

To solve for u₁'(x) and u₂'(x), we equate the coefficients of like powers of x on both sides.

The equation becomes:

3u₁''(x) + 7u₁'(x) + 3u₂''(x)x⁻² + 7u₂'(x)x⁻² = 0 (for x² terms)

3u₁'(x)\(x^{-4/3}\) + 7u₊(x)\(x^{-4/3}\) = 0 (for \(x^{-4/3}\) terms)

u₂'(x)x⁻² + u₂(x)x⁻² = 1 (for x⁻² term)

Now, we integrate the equations obtained in step 3 to find u₁(x) and u₂(x).

Integrating the first equation, we get:

3u₁'(x) + 7u₁(x) + 3u₂''(x)x⁻² + 7u₂'(x)x⁻² = 0

Integrating the second equation, we have:

\(3u_1(x)x^{-4/3} + 7u_1(x)x^{-4/3} = 0\)

Integrating the third equation, we obtain:

u₂(x)x⁻² = x + C₃

where C₃ is the constant of integration.

Substituting the found values of u₁(x) and u₂(x) back into the particular solution, we obtain the general solution of the differential equation:

\(y(x) = y_c + y_p\)

y(x) = C₁ \(x^{-1/3}\) + C₂ \(x^{-1}\) + u₁(x)\(x^{-1/3}\) + u₂(x)x⁻¹.

This is the general solution of the given differential equation obtained by the method of variation of parameters.

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The length of fh is 16 units.

What are congruent triangles ?

If all three corresponding sides and all three corresponding angles are equal in size, two triangles are said to be congruent.

Given that,

The length of side fg is = 3x - 8

The length of side gh is = 16

The length of side fj is = x

In Δfgh, gh is an angle bisector and ⊥ to fh.

In triangles, fgj and hgf

gj ≅ gj           (common)

∠jgf = ∠jgh  (angle bisector)

fg ≅ gh           (c.p.c.t)

Substitute the values of fg and gh,

3x - 8 = 16

3x = 24

x = 8

now, length of fj = 8

As, fj = hj  (c.p.c.t)

Now, by segmentation addition property

fh = fj + hj

fh = 8 + 8

fh = 16

Therefore, the length of fh is 16 units.

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Answer: 5 (4x + 7)

Step-by-step explanation:Wouldn't it be 5(4x+7). So I guess the sign it's looking for is an addition symbol.

The first derivative of the implicit function is equal to \(y' = \frac{\frac{x^{2}}{\left(\sqrt[4]{\frac{1}{3}\cdot x^{2}}\right)^{3}}-\sqrt[4]{\frac{1}{3}\cdot x^{2}}}{4\cdot x}\).

In this question we have the case of an implicit function, that is, a function where a set of variables is not a function of another function, whose first derivative must be found. First, write the entire expression:

x³ / y³ = 3 · x · y

Second, use differentiation rules:

(3 · x² · y³ - 3 · x³ · y² · y') / y⁶ = 3 · y + 3 · x · y'

Third, expand and simplify the expression:

3 · x² / y³ - 3 · x³ · y' / y⁴ = 3 · y + 3 · x · y'

3 · x · y' + 3 · x³ · y' / y⁴ = 3 · x² / y³ - 3 · y

3 · x · (1 + x² / y⁴) · y' = 3 · (x² / y³ - y)

y' = (x² / y³ - y) / [x · (1 + x² / y⁴)]

Fourth, eliminate the variable y by substitution:

x³ / y³ = 3 · x · y

(1 / 3) · x² = y⁴

\(y = \sqrt[4]{\frac{1}{3}\cdot x^{2}}\)

\(y' = \frac{\frac{x^{2}}{\left(\sqrt[4]{\frac{1}{3}\cdot x^{2}}\right)^{3}}-\sqrt[4]{\frac{1}{3}\cdot x^{2}}}{x\cdot \left(1 + \frac{x^{2}}{\frac{1}{3}\cdot x^{2}}\right)}\)

\(y' = \frac{\frac{x^{2}}{\left(\sqrt[4]{\frac{1}{3}\cdot x^{2}}\right)^{3}}-\sqrt[4]{\frac{1}{3}\cdot x^{2}}}{4\cdot x}\)

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Answer:

-2

Step-by-step explanation:

-2 is greater than -4 because on number line the numbers on the left of zero are negative numbers and the numbers which are very close to zero on left side are greater. So -2 is more close to zero than -4 . Hence answer will be

Answer:

the gym that has 150, will take 2 months and the gym that has 50 will take 6 months

first gym:

150

300

second gym:

50

100

150

200

250

300

Answer: anything that greater than 5

Step-by-step explanation

x + 7 > 12

Subtracting a 7 both sides

x > 5 or greater than 5 ( that's an answer to the graph)

Answer:

(5,∞)

Step-by-step explanation:

The height of the pole is 20 feet (to the nearest foot).

Hence, option (B) is the correct answer.

Given that Marci, who is 5 feet tall, stands so that her lines of sight to the top and bottom of the pole form a 90° angle.

.Let the height of the pole be x feet

Now, using similar triangles, we can say that:

`5/x = x/h`

Where h is the distance between Marci and the pole.

Hence, we can write:

`h^2 = x^2 + 5^2`or`h^2 = x^2 + 25`

Now, using the given data, we know that `h = x + 5`

Thus, substituting h in the second equation, we get:

`(x+5)^2 = x^2 + 25`

Expanding the terms, we get:`

x^2 + 10x + 25 = x^2 + 25`

Simplifying the terms, we get:

`10x = 0x = 0`

Thus, the height of the pole is:

`x = h - 5`

But we know that h = x + 5

Hence, `x = h - 5 = (x+5) - 5 = x`

Therefore, the height of the pole is:x = 20 feet

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Answer:

Step-by-step explanation:

Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes!* See Answer. *Response times may vary by subject and question

The inflection points of the given function are at x = 0 only. The answer is D.

To find the inflection points of the given function, we need to find the second derivative of the function and set it equal to zero.

y = 5x^4 - x^5

y' = 20x^3 - 5x^4

y'' = 60x^2 - 20x^3

Setting y'' = 0, we get:

60x^2 - 20x^3 = 0

20x^2(3 - x) = 0

x = 0 or x = 3

Now, we need to determine whether these values of x correspond to inflection points. To do this, we can examine the sign of the second derivative in the intervals around each value of x.

For x < 0, y'' is positive (since both terms are positive), so there is a local minimum at x = -3.

For 0 < x < 3, y'' is negative (since 60x^2 < 20x^3), so there is a point of inflection at x = 0.

For x > 3, y'' is positive (since both terms are positive), so there is a local minimum at x = 3.

Therefore, the inflection points of the given function are at x = 0 only. The answer is D.

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Answer:

∠1 is 101 ∠4 is 79 and ∠2 is 79

Step-by-step explanation:

This is because ∠1 is a vertical angle of ∠3 so they have the same angle

For ∠4 and ∠2 you would do 101-180 because they are supplementary do they add up to 180