there are 1,560 freshmen and 1,830 sophomores at a prep rally at noon. after 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. find the ratio of freshmen to sophomores at 1 p.m.

To find the equation of the tangent line to the curve y = sec(x) at the point (π/3, 2), we need to take the derivative of y with respect to x.

The derivative of sec(x) is sec(x)tan(x), so at x = π/3, we have: y' = sec(π/3)tan(π/3) = 2√3/3, Now we can use the point-slope form of a line to find the equation of the tangent line: y - y1 = m(x - x1)

where (x1, y1) is the point we're given (in this case, (π/3, 2)) and m is the slope of the tangent line (in this case, 2√3/3).

Plugging in the values, we get:

y - 2 = (2√3/3)(x - π/3)

Simplifying, we get:

y = (2√3/3)x + 2 - 2√3

So the equation of the tangent line to the curve y = sec(x) at the point (π/3, 2) is y = (2√3/3)x + 2 - 2√3.
To find the equation of the tangent line to the curve y = sec(x) at the point (π/3, 2), we need to follow these steps:

Step 1: Find the derivative of the function
The derivative of y = sec(x) is:
y' = sec(x)tan(x)

Step 2: Evaluate the derivative at the given point
At the point (π/3, 2), we can find the value of the derivative:
y'(π/3) = sec(π/3)tan(π/3) = 2*(√3/2) = √3

Step 3: Use the point-slope form of a linear equation
The point-slope form of a linear equation is:
y - y1 = m(x - x1)

Plug in the given point (π/3, 2) and the slope from Step 2 (√3):
y - 2 = √3(x - π/3)

Step 4: Simplify the equation
y - 2 = √3x - π√3/3
y = √3x - π√3/3 + 2

The equation of the tangent line to the curve y = sec(x) at the point (π/3, 2) is y = √3x - π√3/3 + 2.

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If {v1, v2, v3} is an orthonormal basis for a subspace W of a vector space V, then multiplying v3 by a non-zero scalar c does not necessarily give a new orthonormal basis for W.

To see why, consider the dot product of cv3 with itself. If v3 is an orthonormal vector, then its norm is 1, so the dot product of cv3 with itself is (cv3) • (cv3) = c^2(v3 • v3) = c^2(1) = c^2. Thus, cv3 has a norm of |c|, rather than 1, and so it cannot be an element of an orthonormal basis for W.

However, multiplying v3 by a scalar does give a new basis for W. Specifically, {v1, v2, cv3} is a basis for W if c is non-zero, since v1 and v2 are orthogonal to cv3, and any vector in W can be written as a linear combination of these three vectors. But this new basis is not orthonormal, since cv3 has a norm of |c|. To obtain an orthonormal basis from this new basis, we can normalize each vector in the basis by dividing it by its norm.

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The only options that are quadratic functions are:

Option A:  5x² + 15 x = 0

Option C: x² - 4x = 4 x + 7

Option E: 3x² + 5 x - 7 = 0

The general form of expression of quadratic functions is:

y = ax² + bx + c

where:

a, b, and c are numbers with a not equal to zero.

The graph of a quadratic function is a curve called a parabola.

Looking at the given options, it is clear that:

6 x - 1 = 4x + 7 is not a quadratic equation because it has only one degree.

2 x - 1 = 0 is a linear equation and not a quadratic equation

x³ - 2 x² + 1 = 0 is a cubic polynomial as it has three degrees.

Thus, only options A, C and E are quadratic equations with two degrees.

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To identify missing values, you can check if there are any responses with blank or incomplete answers in the survey data.

These responses may indicate that the corresponding data is missing. Alternatively, you can use software tools such as Excel or statistical packages like R or Python to automatically identify missing values.

To identify erroneous values, you can check if there are any responses that are outside the expected range of values or seem to be unlikely or illogical. For example, if the survey asks respondents to rate the taste of the candy on a scale of 1-10, a response of 100 would be considered erroneous. Similarly, if a respondent rates the texture of the candy as "very bad" but also rates the creaminess of filling as "very good", this may indicate an error.

Once you have identified missing or erroneous values, you can decide how to handle them. For missing values, you can either exclude the corresponding responses from the analysis or use imputation techniques to estimate the missing values based on other available data. For erroneous values, you may need to contact the respondents to clarify their responses or exclude the corresponding responses from the analysis.

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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 dimensions of this rectangle are 6 units by 7 units.

The coordinates of the new point is (3, -3).

The length of the line segment with end points A and B is 7 units.

The length of the line segment with end points C and D is 11 units.

In order to determine the dimensions of this rectangle, we would use the distance formula. In Mathematics, the distance between two (2) points that are on a coordinate plane can be calculated by using this formula:

Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]

Distance AB = √[(5 + 1)² + (4 - 4)²]

Distance AB = √[36 + 0]

Distance AB = 6 units

For the width, we have:

Distance AC = √[(-1 + 1)² + (4 + 3)²]

Distance AC = √[0 + 49]

Distance AC = 7 units.

In Mathematics and Geometry, a reflection over the x-axis is modeled by this transformation rule;

(x, y)            →      (-x, y)

New point = (-3, -3) →  (3, -3).

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Complete Question:

The coordinates of the vertices of a rectangle are (-1, 4), (5, 4), (5, -3), and (-1, -3). What are the dimensions of this rectangle?

So the probability that Joe will not receive a text in the next 10 minutes is approximately 0.865. So the probability that the next text arrives between 9:07 and 9:10 p.m. is approximately 0.149. So the probability that a text arrives before 9:03 p.m. is approximately 0.393. So the probability that no text will arrive for 7 minutes, given that no text has arrived for 5 minutes, is approximately 0.394.

(a) To find the probability that Joe will not receive a text in the next 10 minutes, we can use the cumulative distribution function (CDF) of the exponential distribution. The CDF gives the probability that the time until the next text is less than or equal to a given time t. The CDF of an exponential distribution with mean 5 minutes is:

[tex]F(t) = 1 - e^{(-t/5)}[/tex]

To find the probability that Joe will not receive a text in the next 10 minutes, we need to find F(10):

[tex]F(10) = 1 - e^{(-10/5)}[/tex]

[tex]= 1 - e^{(-2)}[/tex]

≈ 0.865

(b) To find the probability that the next text arrives between 9:07 and 9:10 p.m., we need to find the probability that the time until the next text is between 7 and 10 minutes. We can use the CDF again to find this probability:

P(7 < X < 10) = F(10) - F(7)

[tex]= (1 - e^{(-10/5)}) - (1 - e^{(-7/5)})[/tex]

[tex]= e^{(-7/5)} - e^{(-2)}[/tex]

≈ 0.149

(c) To find the probability that a text arrives before 9:03 p.m., we need to find the probability that the time until the next text is less than 3 minutes. We can use the CDF again to find this probability:

P(X < 3) = F(3)

[tex]= 1 - e^{(-3/5)}[/tex]

≈ 0.393

(d) To find the probability that no text will arrive for 7 minutes, given that no text has arrived for 5 minutes, we can use the memoryless property of the exponential distribution. The memoryless property states that the conditional distribution of the time until the next text, given that no text has arrived in the first 5 minutes, is the same as the original distribution. In other words, the fact that no text has arrived in the first 5 minutes does not affect the probability of a text arriving in the next 7 minutes.

Therefore, the probability that no text will arrive for 7 minutes, given that no text has arrived for 5 minutes, is the same as the probability that no text will arrive for 7 minutes starting from scratch. This is the probability that the time until the next text is greater than 7 minutes. Using the CDF of the exponential distribution, we can calculate:

P(X > 7) = 1 - F(7)

[tex]= 1 - (1 - e^{(-7/5)})[/tex]

= [tex]e^{(-7/5)}[/tex]

≈ 0.394

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The value of the indicated angle is 128⁰.

The value of the indicated angle is calculated applying intersecting chord theorem as shown below.

The tangent angle formed outside the circumference is equal to the angle formed by the intersection of the chords at the center.

Angle adjacent the indicated angle = 52⁰ (intersecting chord theorem)

The value of the indicated angle is calculated as follows;

52⁰ + ? = 180 ( sum of angles on a straight line )

? = 180 - 52⁰

? = 128⁰

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The probability that both professors are chemistry professors is 0.467.

The cost of a Statistics book is a continuous variable because it can take any value within a given range, including decimals, and is not limited to specific, separate values.

For the probability question, there are 10 professors in total (3 statistics + 7 chemistry), and the organization needs 2 advisors. The probability of choosing two chemistry professors can be calculated using the formula: P(Chemistry) = (Number of Chemistry professors) / (Total Number of professors).

First, find the probability of selecting a chemistry professor for the first advisor:
P(First Chemistry) = 7/10.

Next, find the probability of selecting a chemistry professor for the second advisor, considering one chemistry professor has already been selected:
P(Second Chemistry) = 6/9.

Now, multiply both probabilities together to get the probability of selecting two chemistry professors:
P(Both Chemistry) = P(First Chemistry) × P(Second Chemistry) = (7/10) × (6/9) = 0.467.

So, the probability that both professors are chemistry professors is 0.467.

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William owes a total of $10,142.84.  To calculate this, we first need to determine the total interest paid over the 3-year period.

Using the formula A = P(1 + r/n)^(nt), where:
A = the final amount owed
P = the principal amount borrowed (in this case, $9000)
r = the interest rate (4.5%)
n = the number of times the interest is compounded per year (annually in this case)
t = the number of years (3 in this case)

We can plug in the values:
A = 9000(1 + 0.045/1)^(1*3)
A = 9000(1 + 0.045)^3
A = 9000(1.1412)
A = 10,270.88

This gives us the total amount owed after 3 years. But we only want to know the principal plus interest, so we subtract the original amount borrowed:

10,270.88 - 9000 = 1270.88

Therefore, William owes a total of $10,142.84 (rounded to the nearest cent).

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Therefore, the probability that Sayan can go on the trip without having to replace the hard drive during the trip is approximately 0.868.

Since the number of hours that a computer hard drive can run before it conks off is exponentially distributed with an average value of 5 years, we can use the following exponential probability density function:

f(x) = (1/μ) * exp(-x/μ)

where x is the number of hours, μ is the mean (or average) number of hours, and exp() is the exponential function.

In this case, μ = 5 years * 12 months/year = 60 months. We want to find the probability that the hard drive will not fail during an 8 month trip, given that it has already been in use for 3 years (or 36 months).

Let X be the number of months the hard drive lasts. Then, X is exponentially distributed with mean μ = 60 months. We want to find P(X > 8 + 36 | X > 36).

Using the memoryless property of the exponential distribution, we have:

P(X > 8 + 36 | X > 36) = P(X > 8)

= exp(-8/60)

≈ 0.868

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The distance between the points (2, 3) and (-1, -4) is equal to √58 units.

In Mathematics and Geometry, the distance between two (2) end points that are on a coordinate plane can be calculated by using the following mathematical equation:

Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]

Where:

x and y represent the data points (coordinates) on a cartesian coordinate.

By substituting the given end points into the distance formula, we have the following;

Distance = √[(-1 - 2)² + (-4 - 3)²]

Distance = √[(-3)² + (-7)²]

Distance = √[9 + 49]

Distance = √58 units.

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Shift from the starting demand curve to inward.

If there was a demand factor shift that increased the world food demand in 2007-2008, then the world food demand curve would shift outward, to the right of the original demand curve. This means that at any given price, the quantity of food demanded would be higher than before the shift.

This is because an increase in demand factors, such as population growth or changes in consumer preferences, would cause consumers to demand more food at each price level. As a result, the entire demand curve would shift to the right, indicating a higher quantity demanded at every price level.

On the other hand, if there was a demand factor shift that decreased the world food demand, then the demand curve would shift inward, to the left of the original demand curve. This would indicate a lower quantity demanded at every price level.

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The probability that a randomly selected customer orders an appetizer or dessert or both is 0.57 or 57%.

The probability a randomly selected customer orders an appetizer or dessert or both is determined using the formula for the probability of the union of two events:

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

where:

Data given:

P(A) = 0.52,

P(B) = 0.32,

P(A and B) = 0.27.

Solving for P(A or B):

P(A or B) = 0.52 + 0.32 - 0.27

P(A or B) = 0.57

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The probability that the second card is a spade given the first one is a spade is 12/51.

To find the probability that the second card drawn is a spade given that the first one is a spade, we need to use conditional probability.

Let's first find the probability of drawing a spade on the first draw. Since there are 13 spades in a standard deck of 52 cards, the probability of drawing a spade on the first draw is 13/52 or 1/4.

Now, since we are drawing without replacement, there are only 51 cards left in the deck for the second draw and only 12 spades left. So, the probability of drawing a spade on the second draw given that the first one was a spade is 12/51.

Therefore, the probability that the second card is a spade given that the first one is a spade is 12/51 or approximately 0.235.

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Point E to point B and point B to Point E matches will you move between if you move 3 units left and 4 units down

To determine which two points you will move between if you move 3 units left and 4 units down, first let's look at the coordinates of each point:

Point A: (8, 3)

Point B: (9, 10)

Point C: (3, 9)

Point D: (2, 1)

Point E: (6, 6)

Now, let's move 3 units left and 4 units down for each point and find the new coordinates:

Point A: (8 - 3, 3 - 4) = (5, -1)

Point B: (9 - 3, 10 - 4) = (6, 6)

Point C: (3 - 3, 9 - 4) = (0, 5)

Point D: (2 - 3, 1 - 4) = (-1, -3)

Point E: (6 - 3, 6 - 4) = (3, 2)

Point A to Point E: (5, -1) to (3, 2) - No match

Point B to Point C: (6, 6) to (0, 5) - No match

Point B to Point E: (6, 6) to (3, 2) - This matches the movement between Point B and Point E

Point E to Point B: (3, 2) to (6, 6) - This matches the movement between Point E and Point B

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2, -8

The solution set for an equation is all of the values that make the equation true.

Factoring

One way to solve many quadratic equations, especially those with a leading coefficient of 1, is factoring. Remember that quadratic functions are written in the form of ax² + bx + c. This means that the b-value is 6 and the c-value is -16.

In order to factor the equation above, we need to find 2 factors that add to b and multiply to c. In this case, these 2 factors are -2 and 8. Now, plug these values into the form (x+A) * (x+B) = 0.

Now it is easy to see that if x = 2, then the equation will be true.

Additionally, if x = -8, then the equation will be true. The factored form of the quadratic lets us easily see that the solution set is 2 and -8.

Solution Sets

The solution set to an equation is the values that make the equation true. Since this equation is a quadratic set equal to zero, the solution set is also the zeros of the function. Zeros are points where the function crosses the x-axis (aka where y = 0). For quadratics, there can be 0, 1, or 2 real solutions. In this case, the function crosses the x-axis in 2 separate locations, so the solution set has 2 answers.

The perimeter of the bottom of the storage container is,

⇒ 24 feet

We have to given that;

The storage container has a length of 7 feet and a width of 5 feet.

Hence, we get;

The perimeter of the bottom of the storage container is,

⇒ 2 (7 + 5)

⇒ 2 × 12

⇒ 24 feet

Thus, The perimeter of the bottom of the storage container is,

⇒ 24 feet

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volume (v) = 105

volume (v) = 105

volume (v) = 105 formula to find volume

volume (v) = 105 formula to find volumevolume= v

volume (v) = 105 formula to find volumevolume= vlength = l

volume (v) = 105 formula to find volumevolume= vlength = lwidth = w

volume (v) = 105 formula to find volumevolume= vlength = lwidth = w v = l x w x h

volume (v) = 105 formula to find volumevolume= vlength = lwidth = w v = l x w x hv = 5 x 3 x 7

volume (v) = 105 formula to find volumevolume= vlength = lwidth = w v = l x w x hv = 5 x 3 x 7v = 15 x 7

v = 105

The value of each variable in the given circle are:

a = 65.5°    b = 90°    c = 49°

In order to find the value of each variable in the given circle, recall the following:

Find a:

a = 1/2(131)

a = 65.5°

Find b:

b is an inscribed angel of half of the circle, therefore:

b = 90°

Find c:

c = 180 - 131

c = 49°

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The solution is : the probability, to the nearest 10th of a percent, that both marbles drawn will be green will be 0.053.

Here, we have,

n(r) = 7

n(b) = 8

n ( g) = 5

Total number of marbles = 20

The probability that both marbles drawn will be green will be

The probability of the first being green = 5/20 and the probability that the second marble is green = 4/19.

We will multiply the two together , we have

5/20 x 4/19

= 20/380

= 0.05263157895

Therefore : the probability, to the nearest 10th of a percent, that both marbles drawn will be green will be 0.053

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The coordinates of the reflected triangle are A' = (-3, -4), B' = (1, -4) and C' = (3, -1)

Given that, the graph of ABC has coordinates A(-3,4) B(1,4) and C(3,1) which is reflected across the x-axis,

So,

The rule of the reflection across the x-axis is = (x, y) → (x, -y)

So, the coordinates of the reflected triangle will be,

A' = (-3, -4)

B' = (1, -4)

C' = (3, -1)

The graph is attached.

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However, in some cases, the default location of the click point may be set by default to the top-left corner of the screen or window, which would correspond to the coordinate (0, 0) in a Cartesian coordinate system.

If no coordinates have been assigned to the cursor, the default location of the click point will depend on the program or application being used. In most cases, the default location will be the center or starting position of the screen or window in which the program is running, which could be any location on the screen. Therefore, none of the options provided is necessarily correct.

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To determine the required sample size for a political poll with a 4% margin of error and a 97.5% confidence level, a formula can be used. For this scenario, the sample size required would be approximately 862 respondents.

To calculate the sample size needed for a political poll with a 4% margin of error and a 97.5% confidence level, the following formula can be used:

n = (Z^2 * p * (1-p)) / E^2

Where:

n is the sample size

Z is the Z-score associated with the desired confidence level (in this case, it is 2.24)

p is the expected proportion of support for the candidate (this value is typically unknown, so a conservative estimate of 0.5 is often used to get the maximum sample size)

E is the margin of error

Plugging in the values for this scenario, we get:

n = (2.24^2 * 0.5 * (1-0.5)) / 0.04^2

n ≈ 862

Therefore, the required sample size for this political poll is approximately 862 respondents. This sample size would provide a margin of error of 4% at a 97.5% confidence level, meaning that there is a 97.5% chance that the true proportion of support for the candidate lies within the range of the survey results plus or minus the margin of error.

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Using order of magnitude the capacity a cup would hold would be 250 milliliters of liquid.

Order of magnitude is the relative size of a quantity

To determine if a cup would hold 250 liters of liquid or 250 milliliters of liquid, we need to determine the order of magnitude of the capacity of a cup.

We know that the order of magntude of the capacity of a cup is in the order milliliters since it is small and not in the order of magnitude of liters.

So, therefore, the capacity a cup would hold would be 250 milliliters of liquid.

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[tex]\angle AQO + \angle OQB = 120^\circ$, so $\boxed{\theta = 60^\circ}$[/tex] is the solution.

Let's first draw a diagram to visualize the problem:

Since [tex]$\triangle ABC$[/tex] is isosceles, we have [tex]$\angle BAC = \angle BCA = \theta$[/tex] and [tex]$\angle ABC = 180^\circ - 2\theta$[/tex]. Let O be the center of the circle, and let D and E be the points of tangency of the tangents through A and C, respectively.

Since [tex]$AD$[/tex] and [tex]$CE$[/tex] are tangents to the circle, we have [tex]$OD \perp AD$[/tex] and [tex]$OE \perp CE$[/tex]. Since [tex]$AO$[/tex] and [tex]$CO$[/tex] are radii of the circle, we have [tex]$AO \perp BC$[/tex] and [tex]$CO \perp AB$[/tex].

Therefore, [tex]$OD \parallel CO$[/tex] and [tex]$OE \parallel AO$[/tex]. It follows that [tex]$\angle DOB = \angle COB = \theta$[/tex], and [tex]$\angle EOA = \angle AOB = \theta$[/tex].

Thus, [tex]$\angle DOA = \angle EOC = \theta$[/tex], and [tex]$\angle AOE = \angle DOC = \theta$[/tex].

Now, consider the quadrilateral [tex]$AODE$[/tex]. Since [tex]$\angle AOE = \angle DOA = \theta$[/tex], we have [tex]$\angle OAE = \angle ODE = 90^\circ - \theta$[/tex]. Similarly, [tex]$\angle OCE = \angle OED = 90^\circ - \theta$[/tex]. Since [tex]$\triangle ABC$[/tex] is isosceles, we have [tex]$AB = BC$[/tex], so [tex]$AD = EC$[/tex].

It follows that [tex]$\triangle AOD \cong \triangle CEO$[/tex] by SAS congruence.

Therefore, [tex]$AO = CO$[/tex], so [tex]$\triangle AOB$[/tex] and [tex]$\triangle COB$[/tex] are congruent by SAS congruence.

Thus, [tex]$AB = BC$[/tex], so [tex]$\triangle ABC$[/tex] is equilateral.

Therefore, [tex]$\theta = 60^\circ$[/tex].

Finally, since [tex]$\angle AOB = \theta = 60^\circ$[/tex], we have [tex]$\frac{1}{2}\angle APB = \theta = 60^\circ$[/tex], so [tex]$\angle APB = 120^\circ$[/tex].

Therefore, [tex]$\angle APQ = \angle BPQ = \frac{1}{2}(180^\circ - \angle APB) = 30^\circ$[/tex]. Since [tex]$\triangle APQ$[/tex] is isosceles, we have [tex]$\angle QAP = \angle QPA = \frac{1}{2}(180^\circ - \angle APQ) = 75^\circ$[/tex].

Therefore, [tex]$\angle AQP = \angle BQP = 105^\circ$[/tex], so [tex]$\angle AQB = 2\angle BQP = 210^\circ$[/tex].

Thus, [tex]$\angle AQO = \frac{1}{2}(360^\circ - \angle AQB) = 75^\circ$[/tex], and [tex]$\angle OQB = \frac{1}{2}(180^\circ - \angle AQB) = 45^\circ$[/tex].

Therefore, [tex]\angle AQO + \angle OQB = 120^\circ$, so $\boxed{\theta = 60^\circ}$[/tex] is the solution.

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