help! i need the answers for gina wilson all things algebra area of shaded regions unit 8 homework 4 the sheet is attached below

The distance between Algiers and Barcelona is approximately 517.7 miles.

To find the distance between Algiers and Barcelona, we need to find the length of the curve that runs along the surface of the Earth between these two points.

We can start by calculating the angle between the equator and each city. The angle between Algiers and the equator is 90 - 36.8 = 53.2 degrees, and the angle between Barcelona and the equator is 90 - 41.4 = 48.6 degrees.

Since Algiers and Barcelona lie on the same line of longitude, the difference in longitude between the two cities is 0 degrees.

Using the Law of Cosines, we can calculate the distance between the two cities as follows:

distance = sqrt( (3960 miles)^2 + (3960 miles)^2 - 2 * (3960 miles) * (3960 miles) * cos(53.2 - 48.6) )

This simplifies to:

distance = 517.7 miles

Therefore, the distance between Algiers and Barcelona is approximately 517.7 miles.

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Answer:

Yep!

Step-by-step explanation:

Answer: rational

Step-by-step explanation:

The chi-square statistic is used to determine whether there are differences between two nominally scaled variables. Chi-Square is a statistical test used to compare two nominal variables to determine whether they differ.

The Chi-Square test is used to test for differences between two groups. When you want to compare groups or examine the relationship between two variables, this test is useful. The chi-square test is a method of statistical inference that can be used to compare observed frequencies with expected frequencies, allowing us to determine whether there is a meaningful difference between them. When the p-value obtained from a chi-square test is less than 0.05, it is generally considered statistically significant.

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Answer:

Step-by-step explanation:

Multiply the numerator and denominator of rational numbers by 5

\(\dfrac{-1}{2}=\dfrac{(-1)*5}{2*5}=\dfrac{-5}{10}\\\\\\\dfrac{1}{2}=\dfrac{1*5}{2*5}=\dfrac{5}{10}\\\\\\In\ between \ rational \ numbers \\\\\dfrac{-4}{10} ; \dfrac{-3}{10}; \dfrac{-2}{10}; \dfrac{-1}{10}\)

Standard error for sample proportion = √{[p₁/p₂(1-p₁/p₂)]/2}

What means standard error?

Standard error is a statistical statement that express the accuracy of an estimation resulting from survey study. It indicates how much a sample means is differ from actual population mean.

How to calculate standard error?

We calculate standard error for the two-survey group by using the following formula.

√[p(1-p)/n]

p is the proportion of freshmen living on campus to the freshmen living off campus

p = p₁/p₂

According to the given question:

p₁ = freshmen living on campus

p₂= freshmen living off campus

n= number of survey group

n=2

hence, standard error =√{[p₁/p₂(1-p₁/p₂)]/2}

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Answer:

Reflection

Step-by-step explanation:

With a reflection, a shape is mirrored across a line. So, while its position is changed, the points end up getting flipped too. In this problem, you can imagine a diagonal line goes right through the center of the two shapes shown (the original and its mirror). The line would be equidistant from L and G; it would be equidistant from B and D; it would also be equidistant from U and O. If you were to mirror B, U, and L across this line, you would get your reflection points D, O, and G.

The shape BUL is mirrored over a diagonal line to get DOG, which means it is a reflection

Answer:

I’m pretty sure u would spin it 4 times

Step-by-step explanation:

I hope that helps check with someone else tho ❤️❤️

Answer:70 minutes

Step-by-step explanation:this is because since 1 min =6 cartons .And u are figuring out for 420 cartons ,you think what is 420 divide 6 which is 70 .So it means 70.

1 min = 6 cartons

x min = 420 cartons

do 420 times 1 it gives u 420 then divide by 6.

Answer: ABCD is a quadrilateral

To prove : ∠AOB=

2

1

​

(∠C+∠D)

AO and BO is bisector of A and B

∠1=∠2∠3=∠4...(1)

∠A+∠B+∠C+∠D=360

(Angle sum property)

2

1

​

(∠A+∠B+∠C+∠D)=180...(2)

In △AOB

∠1+∠3+∠5=

2

1

​

(∠A+∠B+∠C+∠D)

∠1+∠3+∠5=∠1+∠3+

2

1

​

(∠C+∠D)

∠AOB=

2

1

​

(∠C+∠D)

​

Explanation: In a quadrilateral ABCD. AO and BO are bisectors of angle A and angle B respectively. Prove that ∠AOB=

2

1

​

{∠C+∠D}.

Answer:

The roots of the equation, 3.1·x³ - 2.4·x²+ 6·x - 3 = 4·x² + 3·x + 2,  are;

x = 1.986, x = 0.0392 - 0.9·i, x = 0.0392 + 0.9·i

Step-by-step explanation:

The given equation is 3.1·x³ - 2.4·x²+ 6·x - 3 = 4·x² + 3·x + 2

Which gives;

3.1·x³ - 2.4·x²+ 6·x - 3 - 4·x² - 3·x - 2 = 0

3.1·x³ - 6.4·x²+ 3·x  - 5 = 0

Factorizing online, we get;

3.1·x³ - 6.4·x²+ 6·x + 3·x - 5 = 3.1·(x - 1.986)·(x² - 0.0784·x + 0.812) = 0

Therefore, the possible solutions are;

x - 1.986= 0 or x² - 0.0784·x + 0.812 = 0

The roots of the equation are x² - 0.0784·x + 0.812 = 0 are;

x = 0.0392 - 0.9·i, x = 0.0392 + 0.9·i

Therefore, the roots of the equation, 3.1·x³ - 2.4·x²+ 6·x - 3 = 4·x² + 3·x + 2,  are;

x = 1.986, x = 0.0392 - 0.9·i, x = 0.0392 + 0.9·i.

This is going to take a while: get ready...

I like two-columns, so...

Statements_______________|_____Reasons___

l||m                                                        Given

angle 1 congruent angle 7                   Given

angle 3 congruent angle 5                 corresponding                                                                            alternate angles

a||b                                                       CPCTC

I usually do proofs for triangles, not parallel lines, but I hope this helped!

The simplified expression is (10/(6y)) × x⁴ which is the law of exponent for this expression.

What is the law of exponent?

The law of exponents, also known as the exponent rules or laws of indices, are mathematical rules that simplify expressions involving exponents. In this case, we have the expression:

(10x⁸y⁴)/(-6x⁴y⁵)

To simplify this expression using the laws of exponents, we can use the following rules:

Product rule: a to the power m × aⁿ = a to the power (m+n)

Quotient rule: a to the power m ÷ aⁿ= a to the power (m-n)

Power rule: (a to the power m)ⁿ= a to the power (mⁿ)

First, let's simplify the numerator using the product rule:

10x⁸y⁴= 2 × 5 × x⁸ × y⁴

Next, let's simplify the denominator using the product and quotient rules:

-6x⁴y⁵= -1 × 2 × 3 × x⁴× y⁵

Now, we can rewrite the expression as:

(2 × 5 × x⁸× y⁴)/(-1 × 2 × 3 × x⁴ × y⁵)

Using the quotient rule, we can simplify the expression further:

(2 × 5 × x⁽⁸⁻⁴⁾ × y⁽⁴⁻⁵⁾)/(-1 × 2 × 3)

Simplifying the exponents, we get:

(10x⁴y⁻¹)/(-6)

Finally, we can simplify this expression using the power rule and simplify the negative exponent by moving it to the denominator:

(10/(6y)) × x⁴

Therefore, the simplified expression is (10/(6y)) × x⁴, which is the law of exponent for this expression.

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Answer:

20.19°

Step-by-step explanation:

Find the diagram to the question attached. We are to look for <CAB

Applying sine rule on ΔABC:

\(\frac{4.5}{sin15^0} = \frac{6}{sin<CAB}\)

cross multiply

\(4.5 * sin<CAB =6 sin15^0\\\\sin<CAB = \frac{6sin15^0}{4.5}\\ \\sin<CAB = \frac{6*0.2588}{4.5}\\ \\sin<CAB = \frac{1.5529}{4.5}\\\\sin<CAB = 0.3451\\\\<CAB = sin^{-1}0.3451\\\)

\(<CAB = 20.19^0\\\)

Hence the angle <CAB is 20.19°

The angle ∠CAB is 20.19° and the distance (d) Ollie is from the entrance to his house when he first activates the sensor is 5.77 m.

Trigonometry deals with the relationship between the sides and angles of a right-angle triangle.

Ollie has installed security lights on the side of his house that is activated by a sensor.

The sensor is located at point C directly above point D.

The area covered by the sensor is shown by the shaded region enclosed by triangle ABC.

The distance from A to B is 4.5 m and the distance from B to C is 6 m.

The angle ACB is 15°.

The ∠CAB is ∠A.

We know that the sine rule

\(\dfrac{a}{\sin A} =\dfrac{b}{\sin B} = \dfrac{c}{\sin C}\)

We have

\(\dfrac{AB}{\sin C} \ \ = \dfrac{BC}{\sin A} = \dfrac{AC}{\sin B}\\\\\\\dfrac{4.5}{\sin 15^o} = \dfrac{6}{\sin A} = \dfrac{b}{\sin B}\)

From the first two, we have

\(\dfrac{4.5}{\sin 15^o} = \dfrac{6}{\sin A} \\\\\sin A = \dfrac{6}{4.5} \times \sin 15^o\\\\A \ \ \ \ = 20.19^o\)

We know that the sum of the angle is 180 degrees. Then we have

    ∠A + ∠B + ∠C = 180°

20.19° + ∠B + 15° = 180°

                      ∠B = 144.81°

Then the value of b will be

\(\dfrac{4.5}{\sin 15^o} = \dfrac{b}{\sin 144.81^o}\\\\\\b \ \ \ = 10.02\)

Then the distance (d) Ollie is from the entrance to his house when he first activates the sensor will be

d = b × sin 144.81°

d = 10.02 × sin 144.81°

d = 5.77 m

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Answer:

4

Step-by-step explanation:

Point slope Formula: y2-y1/x2-x1

10+2/7-4

=4

Answer:

2

Step-by-step explanation:

30y x 2 = 60y

Answer:

524 divided by n

524 ÷ n

524/n

The interval estimate of the population mean when a random sample is taken of the speed on highway is 59.5<h<64.1. Option C is correct.

The probability or the chances of error while choosing or calculating a sample in a survey is called the margin of error.

The interval estimate of the population mean can be found out using the following formula:

\(\overline X\pm MOE\)

Here, \(\overline X\) is the average mean, and the MOE is the margin of error.

On a highway, a random sample is taken of the speed, in mph, of 50 cars.

Put the values in the above formula as,

\(61.8\pm 2.3\\59.5,641\)

Thus, the interval estimate of the population mean

\(59.5 < h < 61.8\)

Hence, the interval estimate of the population mean when a random sample is taken of the speed on highway is 59.5<h<64.1. Option C is correct.

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The following: (a) λ² + (1/4) = 0. (b)  λ = ±√(-1/4). (c)\(e^{j\frac{\pi}{4n}} \quad \text{and} \quad e^{-j\frac{\pi}{4n}}\). (d) Homogeneous response:\(y_h[n] = C_1 \times e^{\frac{j\pi}{4n}} + C_2 \times e^{-\frac{j\pi}{4n}}\), (e) \(x[n] = (1/2)^n \times u[n]\) as the input, (f) input x[n] (g) \(y[n] = y_h[n] + y_p[n].\)

(a) The characteristic polynomial is obtained by assuming a solution of the form \(y[n] = y_h[n] + y_p[n].\) and substituting it into the difference equation.

(b) To find the characteristic roots, we solve the characteristic polynomial for λ. The roots will be complex conjugates with a negative real part, as indicated by the presence of the square root of a negative number.

(c) The characteristic modes arise from the complex roots and are of the form e^(jωn) and e^(-jωn), where ω is the angle of the roots in polar form.

(d) The homogeneous response is the general solution to the difference equation with the initial conditions set to zero, and it contains the characteristic modes.

(e) The impulse response is found by setting the initial conditions y[-1] and y[-2] to zero and solving the difference equation with x[n] = (1/2)ⁿ × u[n] as the input.

(f) The particular response is the solution to the difference equation with the given input x[n], which can be found using appropriate methods like undetermined coefficients or convolution.

(g) The total response is the sum of the homogeneous and particular responses, which gives the complete output of the system for the given input and initial conditions.

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The 95% confidence interval for the true population mean, based on a sample with a sample size (n) of 225, a sample mean (X) of 8.5, and a sample standard deviation (σ) of 45, is (2.62, 14.38).

To calculate the confidence interval, we can use the formula:

Confidence interval = X ± Z * (σ/√n)

where X is the sample mean, Z is the critical value for the desired level of confidence (in this case, 95%), σ is the sample standard deviation, and n is the sample size.

The critical value Z can be obtained from a standard normal distribution table or calculated using statistical software. For a 95% confidence level, the Z-value is approximately 1.96.

Plugging in the values into the formula, we get:

Confidence interval = 8.5 ± 1.96 * (45/√225)

                 = 8.5 ± 1.96 * (45/15)

                 = 8.5 ± 1.96 * 3

Calculating the upper and lower bounds of the confidence interval:

Upper bound = 8.5 + 1.96 * 3

          = 8.5 + 5.88

          = 14.38

Lower bound = 8.5 - 1.96 * 3

          = 8.5 - 5.88

          = 2.62

Therefore, the 95% confidence interval for the true population mean is (2.62, 14.38).

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The value of the integral is 343π/3 by changing to polar coordinates. √(49 − x2 − y2) dA where r = (x, y) | x2 y2 ≤ 49, x ≥ 0

It is defined as the mathematical calculation by which we can sum up all the smaller parts into a unit.

We have the integral:

\(\int\limits \int\limits_R {\sqrt{49-x^2-y^2}} \, dA\)

Where,  r = (x, y) | x2 y2 ≤ 49, x ≥ 0

The polar coordinate will be:

x = rcosθ

y = rsinθ

Where x²+y²= r²

Put the value of x and y in the integral, and the limits will be:

r²≤49 or 0≤r≤7, -π/2≤θ≤π/2  ( since x ≥0)

dA = rdrdθ

\(\int\limits \int\limits_R {\sqrt{49-x^2-y^2}} \, dA = \int\limits^{\dfrac{\pi}{2}}_{\dfrac{-\pi}{2}} \int\limits^7_0 {\sqrt{49-r^2]} \, rdrd\theta\)

After solving the double integration, we will get:

\(\int\limits \int\limits_R {\sqrt{49-x^2-y^2}} \, dA = \dfrac{343}{3} \pi\)

Thus, the value of the integral is 343π/3 by changing to polar coordinates. √(49 − x2 − y2) dA where r = (x, y) | x2 y2 ≤ 49, x ≥ 0

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The number of pheasants in 2012 was 1832. The number of pheasants in 2016 was 422. The minimum viable density is 100 pheasants per hectare.

So, we need to find out the year in which the population will fall below 100 pheasants per hectare.

First, we need to find out the rate of decline per year which can be calculated as follows: Rate of decline per year = (1832 - 422) / (2016 - 2012)= 1410 / 4= 352.5 per yearNow, let the number of pheasants per hectare be y after t years from 2012.

Then, y = 1832 - 352.5t (as we know the rate of decline per year)We need to find the year in which the population will fall below 100 pheasants per hectare. So, we need to solve the equation:1832 - 352.5t = 100⇒ 352.5t = 1732⇒ t = 1732 / 352.5⇒ t = 4.91 Therefore, the population will fall below the minimum viable density of 100 pheasants per hectare after 4.91 years from 2012, which is around 2017 or 2018.

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Because the temperature is greater, the relative humidity of the atmosphere is higher.

Water vapor is also measured by relative humidity (RH), which is stated as a percentage but RELATIVE to the air's temperature. In plenty of other terms, it is a comparison between the quantity of water vapor that is really present in the air and the maximum amount water vapor that is possible for the air at the current temperature.

By deducting the temperature just on wet-bulb thermometer from of the temperature just on dry-bulb thermometers and utilizing a relative humidity chart, one may determine the relative humidity.

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Answer:

Step-by-step explanation:

d=m/V

7.8=15/V

V=15/7.8

≈1.92 cm³

Answer:

x=20 degrees

Step-by-step explanation:

So, lets recall some things.

First off, opposite angles are the same.

So this would make x equal 20.

But lets test this.

When you look at a X, or any other sort of two lines that cross each other, can you ever think of a time when one side is not the same as the other? No.

This is because when you change one angle, since the lines are crossed together, it will change the opposite angle. This change will always be constant as long as the two lines connect at the same point

So if you changed a 10 degree angle of two lines, so that it was 30, this would change for the other line.

I attached a image below.

I hope this explains this to you and helps! :D

==================================================================

Respuesta:

x = 20 grados

Explicación paso a paso:

Entonces, recordemos algunas cosas.

En primer lugar, los ángulos opuestos son iguales.

Entonces esto haría que x sea igual a 20.

Pero probemos esto.

Cuando miras una X, o cualquier otro tipo de dos líneas que se cruzan, ¿puedes pensar alguna vez en un momento en el que un lado no sea igual que el otro? No.

Esto se debe a que cuando cambia un ángulo, dado que las líneas se cruzan, cambiará el ángulo opuesto. Este cambio siempre será constante siempre que las dos líneas se conecten en el mismo punto

Entonces, si cambia un ángulo de 10 grados de dos líneas, de modo que sea 30, esto cambiaría para la otra línea.

Adjunto una imagen a continuación.

¡Espero que esto te explique esto y te ayude! :D

Answer:

angle of elevation is at top and depresseion on bottom.

Step-by-step explanation:

far apart and height is calculated with more information

A) The steady-state error for transfer function (20s² + 2s + 36) with reference r(t) = 1 and K = 20 is -35.

B) The steady-state error for transfer function (6s² + 22) with reference r(t) = 1 and K = 20 is -21.

A) Transfer Function: G(s) = (20s² + 2s + 36)

To calculate the steady-state error, we need to evaluate the transfer function G(s) at s = 0.

So let's substitute s = 0 into the transfer function:

G(0) = (20(0)²+ 2(0) + 36)

= 0 + 0 + 36

= 36

The steady-state value of the output for reference r(t) = 1 and K = 20 is 36.

Steady-State Error = r(t) - y(t)

= 1 - 36

= -35

B) Transfer Function: G(s) = (6s² + 22)

Let's evaluate the transfer function G(s) at s = 0:

G(0) = (6(0)² + 22)

= 0 + 22

= 22

The steady-state value of the output for reference r(t) = 1 and K = 20 is 22.

Therefore, the steady-state error is:

Steady-State Error = r(t) - y(t)

= 1 - 22

= -21

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Null hypothesis: The probability of at least two people in each classroom of 25 students sharing a birthday is less than or equal to 40%.

Alternative hypothesis: The probability of at least two people in each classroom of 25 students sharing a birthday is greater than 40%.

Simulation Results:

Simulation 1: 8 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 2: 9 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 3: 7 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 4: 11 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 5: 7 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 6: 6 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 7: 8 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 8: 6 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 9: 10 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 10: 8 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 11: 7 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 12: 11 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 13: 8 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 14: 7 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 15: 10 out of 16 classrooms had at least 2 students sharing a birthday.

Simulation 16: 10 out of 16 classrooms had at least 2 students sharing a birthday.

Proportion of classrooms with at least 2 students sharing a birthday = (8+9+7+11+7+6+8+6+10+8+7+11+8+7+10+10)/16 = 0.7125

The mean number of classrooms with at least 2 students sharing a birthday is 0.7125. We can use this value to calculate the t-statistic.

t-statistic = (0.7125 - 0.4) / (sqrt(0.4 * 0.6 / 16)) = 5.766

Using a one-tailed t-test with 15 degrees of freedom and an alpha level of 0.05, the critical t-value is 1.753. Since the calculated t-statistic is greater than the critical t-value, we can reject the null hypothesis.

The p-value is very small (less than 0.0001), indicating strong evidence against the null hypothesis.

Therefore, we can conclude that there is sufficient evidence to suggest that the probability of at least two people in each classroom of 25 students sharing a birthday is greater than 40%.

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