Colin predicted whether he got answers right or wrong in his 50 question exam.
He identified the 12 questions he thought he got wrong.
It turns out that Colin got 5 questions right that he thought he got wrong.
Colin also got a total score of 42 out of 50 in the test.
What is the percentage accuracy he had with predicting his scores?

The factors to the given expression are -1(x+3)(x-8)

The x-intercepts of the quadratic relationship are -3, 8. When we write an expression in its factors and multiplying those factors gives us the original expression, then this process is known as factorization.

How do we factorize the given expression?

We equate the given expression to f(x)

   (-[tex]x^{2}[/tex] + 5x + 24) = f(x)

⇒ -1([tex]x^{2}[/tex] - 5x - 24) = f(x)

⇒ -1([tex]x^{2}[/tex] - (8-3)x - 24) = f(x)

⇒ -1([tex]x^{2}[/tex] + 3x - 8x -24) = f(x)

⇒ -1(x(x+3) -8(x+3)) = f(x)

⇒ -1(x+3)(x-8) = f(x)

∴The factor to the given expression is -1(x+3)(x-8)

How do we find the x-intercepts?

We equate f(x) = 0 to find the x-intercepts.

⇒ -1(x+3)(x-8) = 0

⇒ (x+3)(x-8) = 0

The roots of the above equation are x-intercepts.

Therefore, the x-intercepts occur where x = -3 and x = 8

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The complete question is "Factor the expression (-x^2 + 5x + 24.) and use the factors to find the x-intercepts of the quadratic relationship it represents.

Type the correct answer in each box, starting with the intercept with the lower value.

The x-intercepts occur where x =

and x = "

Answer:

A. 4^2 - 4(2)(26) = 16 - 208 = -192

B. This equation has no real roots.

C. (-4 + √-192)/(2×2) = (-4 + 8i√3)/4

= -1 + (2√3)i

Answer:

21

Step-by-step explanation:

35% of 60=60% of 35

10% of 35=3.5

3.5*6=21

a. The nth term of the sequence is  11 - 3n.

b. The nth term of the sequence is  14- 5n.

The sequence is an arithmetic progression. Therefore, the expression for the sequence can be represented as follows:

nth term = a + (n + 1)d

where

Therefore,

a.

a = 8

d = 11 - 8 = - 3

Therefore,

nth term = 8 + (n - 1)-3

nth term = 8  - 3n + 3

nth term = 11 - 3n

b.

a = 19

d = 14 - 19 = -5

nth term = 19 + (n - 1)-5

nth term = 19 - 5n - 5

nth term =  14 - 5n

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The time taken by the train to pass the man is 54 seconds

To find the time taken by the train to pass a man standing near the railway line, we need to convert the train's speed to meters per second and then use the formula time = distance/speed.

1. Convert the speed of the train from km/hr to m/s: 40 km/hr * (1000 m/km) / (3600 s/hr) = 40000/3600 = 40/3.6 = 10/0.9 = 100/9 m/s.

2. Now, use the formula: time = distance/speed. The distance is the length of the train (600 m) and the speed is 100/9 m/s.

time = 600 m / (100/9 m/s) = 600 * 9 / 100 = 54 seconds.

Therefore, the time taken by the train to pass the man is 54 seconds (Option A).

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The balance point of the data, given the number of hours rehearsed and the score would be (5, 50).

The balance point on the graph is simply the average of the x vertices and the y vertices.

The average of the x vertices is:

= ( 1 + 3 + 4 + 8 + 9 )  / 5

= 25 / 5

= 5

The average of the y vertices is:

= ( 30 + 50 + 20 + 90 + 60 ) / 5

= 250 / 5

= 50

This then means that the balance point would be ( 5, 50 ).

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The slope of the line given above which passes through points (-2, -1) and (0, -2) is calculated as: m = -1/2.

To find the slope of the line given above, apply the slope formula, then use the coordinates of any two points on the line to calculate the slope.

Slope of a line (m) = change in y / change in x = y2 - y1 / x2 - x1

We have:

(-2, -1) = (x1, y1)

(0, -2) = (x2, y2)

Plug in the values:

Slope (m) = (-2 -(-1)) / (0 - (-2))

m = -1 / 2

Slope of the line (m) = -1/2

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The solution to the system is [tex]$\$\left(x_{-} 1, x_{-} 2\right)=(4,-3) \$$[/tex].

To solve the system, we can use the method of elimination or Gaussian elimination.

We start by writing the system in augmented matrix form:

[tex]$$\left[\begin{array}{cc|c}2 & 4 & -4 \\5 & 7 & 11\end{array}\right]$$[/tex]

We can eliminate the [tex]$\$ x_{-} 1 \$$[/tex] variable from the second equation by subtracting 5 times the first equation from the second:

[tex]$$\left[\begin{array}{cc|c}2 & 4 & -4 \\5-5(2) & 7-5(4) & 11-5(-4)\end{array}\right] \Rightarrow\left[\begin{array}{cc|c}2 & 4 & -4 \\-3 & -13 & 31\end{array}\right]$$[/tex]

Next, we can eliminate the [tex]$\$ x_{-} 2 \$[/tex]$ variable from the first equation by subtracting twice the second equation from the first:

[tex]$$\left[\begin{array}{cc|c}2-2(-13) & 4-2(7) & -4-2(31) \\-3 & -13 & 31\end{array}\right] \Rightarrow\left[\begin{array}{cc|c}28 & -10 & -66 \\-3 & -13 & 31\end{array}\right]$$[/tex]

We can simplify this further by dividing the first row by 2 :

[tex]$$\left[\begin{array}{cc|c}14 & -5 & -33 \\-3 & -13 & 31\end{array}\right]$$[/tex]

Now we can solve for [tex]$\$ x_{-} 2 \$$[/tex] in terms of [tex]$\$ x_{-} 1 \$$[/tex] by multiplying the first equation by 13 and adding it to the second equation:

[tex]$$13(14) x_1-13(5) x_2-13(33)-3(-13) x_1-3(-13) x_2=13(31)-3(14) x_1$$[/tex]

Simplifying:

[tex]$$\begin{aligned}& 169 x_1-91 x_2-429+39 x_1+39 x_2=403 \\& 208 x_1=832 \\& x_1=4\end{aligned}$$[/tex]

Substituting back into the first equation, we get:

[tex]$$2(4)+4 x_2=-4 \Rightarrow x_2=-3$$[/tex]

Therefore, the solution to the system is [tex]$\$\left(x_{-} 1, x_{-} 2\right)=(4,-3) \$$[/tex].

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Answer:

The area of the pool is 10*20 = 200 square meters. Let's assume the width of the path is x. Then the dimensions of the entire region would be (10+2x) by (20+2x). The area of the entire region would be (10+2x)*(20+2x) = 400 + 60x + 4x^2. We know that the area of the pool and the path combined is 1200 square meters. So we can set up the equation as follows:

200 + 1200 = 400 + 60x + 4x^2

Simplifying the equation, we get:

4x^2 + 60x - 1000 = 0

Dividing both sides by 4, we get:

x^2 + 15x - 250 = 0

Factoring the equation, we get:

(x + 25)(x - 10) = 0

x = 10 or x = -25

Since the width of the path can't be negative, the width of the path is 10 meters.

Step-by-step explanation:

Mark Brainliest!!

You need to deposit at least $74 into your bank account to avoid an overdraft fee.

We have,

Start with the initial balance = $120

Subtract the first expense, the electricity bill = $120 - $67 = $53

Subtract the second expense, the night out with friends = $53 - $44 = $9

Subtract the third expense, the gas purchase = $9 - $35 = -$26

Now,

Since the remaining balance is negative, you would receive an overdraft fee if you left it at this amount.

To avoid the overdraft fee, you need to deposit enough money to bring your account balance back to $0 or higher.

To do this, you need to add the absolute value of the negative balance to your desired minimum balance, which in this case is:

$0 = |-26| + $0 = $26

Therefore,

You need to deposit at least $74 into your bank account to avoid an overdraft fee.

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Answer:

69

Step-by-step explanation:

180-45=135

69 is the nearest angle degree.

Given the linear inequality graph, which two statements are true include the following:

B) The graph represents y < −1/3(x) + 5.

D) All points in the blue area are solutions.

In Mathematics and Geometry, the point-slope form of a straight line can be calculated by using the following mathematical equation (formula):

y - y₁ = m(x - x₁)

Where:

First of all, we would determine the slope of this line;

Slope (m) = (y₂ - y₁)/(x₂ - x₁)

Slope (m) = (4 - 7)/(3 + 6)

Slope (m) = -3/9

Slope (m) = -1/3

At data point (3, 4) and a slope of -1/3, a linear equation for this line can be calculated by using the point-slope form as follows:

y - y₁ = m(x - x₁)

y - 4 = -1/3(x - 3)  

y - 4 = x/3 + 1

y = x/3 + 5

y < x/3 + 5

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Missing information:

The question is incomplete and the complete question is shown in the attached picture.

The correct answer is d. The significance level tests for the coefficients are not valid.

Multicollinearity is a statistical term that refers to the presence of high correlation among predictor variables in a regression model. This can cause issues in the model, such as unstable or unreliable coefficients, and can lead to incorrect conclusions about the relationships between the predictors and the response variable.

When multicollinearity is present, the R-squared value of the model can become inflated because the model is able to explain more of the variation in the response variable due to the high correlation among the predictor variables. This can give the impression that the model is a good predictor when in fact it may not be. Additionally, multicollinearity can cause the F-test significance level to be low, indicating that the model is a good fit, even though the individual coefficients may not be statistically significant.

Multicollinearity can cause inflated R-squared values and low F-test significance levels. However, it does not necessarily mean that the model is a poor predictor. The interpretation of coefficients may also be affected by multicollinearity.

However, the most significant issue with multicollinearity is that it can lead to unreliable significance tests for individual coefficients, making it difficult to determine which predictors are contributing significantly to the model.

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We can proceed with finding the specific solution using either method mentioned above.

The order of the differential equation is 2.

Since the auxiliary equation has complex roots (±3i), we know that the general solution to the differential equation will involve sine and cosine functions.

To find the specific solution, we can use the method of undetermined coefficients or variation of parameters. However, we first need to check for any singular points or irregular singular points in the equation.

Since the coefficient of y is a polynomial in x and the coefficient of y" is also a polynomial in x, there are no singular points or irregular singular points in the equation.

Therefore, we can proceed with finding the specific solution using either method mentioned above.

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Answer:

To find the slope of the tangent line to the path of the particle at the point where t = 2, we first need to find the values of x(2) and y(2), as well as their derivatives x'(2) and y'(2).

Using the given parametric functions, we can find:

x(2) = ∫ x'(t) dt = ∫ t sin(t) dt = -t cos(t) + sin(t) + C

where C is the constant of integration.

Since we want x(2), we can evaluate the above expression at t = 2:

x(2) = -2 cos(2) + sin(2) + C

Similarly, we can find:

y(2) = ∫ y'(t) dt = ∫ (5e^(-3t) + 2) dt = (-5/3)e^(-3t) + 2t + C'

where C' is the constant of integration.

Again, since we want y(2), we can evaluate the above expression at t = 2:

y(2) = (-5/3)e^(-6) + 4 + C'

Now we can find the derivatives x'(2) and y'(2) by taking the derivative of x(t) and y(t), respectively, and evaluating them at t = 2:

x'(2) = 2 sin(2) - cos(2)

y'(2) = (5/3)e^(-6)

Therefore, at t = 2, the particle is at the point (x(2), y(2)) = (-2 cos(2) + sin(2) + C, (-5/3)e^(-6) + 4 + C'), and the slope of the tangent line to the path of the particle at this point is given by:

dy/dx = (dy/dt)/(dx/dt) = y'(2)/x'(2)

Substituting the values we found:

dy/dx = [(5/3)e^(-6) + 4 + C']/(2 sin(2) - cos(2))

Since we don't have enough information to find the value of C', we cannot find an exact value for the slope. However, we can simplify the expression by using the trigonometric identities:

sin(2) = 2 sin(1) cos(1)

cos(2) = cos^2(1) - sin^2(1)

where we let t = 1 for simplicity. Then, we can substitute these expressions and simplify:

dy/dx = [(5/3)e^(-6) + 4 + C']/(4 sin(1) cos(1) - cos^2(1) + sin^2(1))

dy/dx = [(5/3)e^(-6) + 4 + C')/(4 sin(1) cos(1) - 1)

Therefore, the slope of the tangent line to the path of the particle at the point where t = 2 is given by the above expression.

Step-by-step explanation:

The slope of the tangent line to the path of the particle at the point where t=2 is approximately 1.55. To find the slope of the tangent line to the path of the particle at the point where t=2,

we need to use the derivatives of x(t) and y(t).

First, we can find the slope of the tangent line by using the formula:

slope = dy/dx = (dy/dt)/(dx/dt)

So, we need to find both dy/dt and dx/dt.

Given that x′(t)=t sin t, we can find dx/dt by taking the derivative of x(t):

dx/dt = x′(t) = t sin t

Given that y′(t)=5e−3t+2, we can find dy/dt by taking the derivative of y(t):

dy/dt = y′(t) = 5e−3t+2

Now, we can find the slope of the tangent line at t=2 by plugging in these values:

slope = (dy/dt)/(dx/dt) = (5e−3t+2)/(t sin t) = (5e−6+2)/(2 sin 2)

Therefore, the slope of the tangent line to the path of the particle at the point where t=2 is approximately 1.55.

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All of the expressions are not a factor of the polynomial function 4x³ − 7x²+x+6

From the question, we have the following parameters that can be used in our computation:

The polynomial function 4x³ − 7x²+x+6

To check the expression that is not a factor, we set the expression to 0, solve for x and calculate the value of the polynomial at this x value

If the result is not zero (0), then it is not a factor of the polynomial

Using the above as a guide, we have the following:

x - 1 gives x = 1

So, we have

4(1)³ − 7(1)² + (1) + 6 = 4

x - 2 gives x = 2

So, we have

4(2)³ − 7(2)² + (2) + 6 = 12

x + 1 gives x = -1

So, we have

4(-1)³ − 7(-1)² + (-1) + 6 = -6

x + 2 gives x = -2

So, we have

4(-2)³ − 7(-2)² + (-2) + 6 = -56

None of the expressions give a solution of 0

Hence, all of the expressions are not a factor of the polynomial function

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Answer:

1) 4 pounds / $5.48 = .73 pounds / dollar

2) 5 pounds / $4.85 = 1.03 pounds / dollar

3) $3.51 / 3 pounds = $1.17 / pound

4) $9.12 / 6 pounds = $1.52 / pound

The time for which plane B will take to refuel plane A is equals to 10 minutes. The two points who satisfy the equation, y = x + 1, are (0, 1), (-1,0). The two points who satisfy the equation, y = -x + 21, are (0,21), (21,0).

We have a fuel left in Plane A = 1 ton

fuel left in Plane B = 21 tons

Fuel transfer rate = 1 ton per minute

In order that for them to have the same amount of fuel, We add up the fuel left in Plane A and Plane B = 21 + 1 = 22 tons. This implies each plane will have fuel of 11 tons. Time that plane B will take to refuel plane A until both planes have the same amount of fuel is calculated by : Plane B will transfer 10 tons of fuel to A.

Plan A has a total of 11 tons. Since, the transfer rate = 1 ton per minute

=> 1 ton will transfer in 1 minute

So, 10 tons fuel will need 10 minutes. Hence, required time value is 10 minutes. Now, The equation for quantity of fuel with respect to time in plane A is, y = x + 1 --(1). If x = 0 => y = 1

and y = 0 => x = -1. So, (0, 1) and (-1,0).

The equation for quantity of fuel with respect to time in plane B is, y = -x + 21 --(2). For it, x = 0 => y = 21 and y= 0 => x = 21. Hence, two points that satisfy the equation(1) and equation(2) are (0, 1), (-1,0) and (0,21), (21,0) respectively.

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Area measures the size of a closed curve in square units. It is the degree of the measure of a two-dimensional region enclosed \by a closed bend. It is solved in square units.

The equation for the areas of diverse plane figures are:

Therefore, for composite figures, which are made up of two or more basic figures, the zone can be found by including the ranges of the person figures. Some of the time, it may be essential to subtract ranges that are numbered twice.

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Using Table B at line 133, the consumers assigned to treatment 1 are those numbered 4, 5, 7, 1, and 8.

The table displaying the treatments in a design with two factors would be: | | Ambient scent | |----------|--------------| | Product density | Seashore (A) | Enclosed space (B) | No scent | | Jam-packed | 2 | 1 | 3 | | Minimally stocked | 4 | $ | 6 |

To randomly assign the 30 subjects to the six treatments, we would first label the subjects from 01 through 30. Then, we would use Table B starting at line 133 to randomly select the appropriate number of subjects for each treatment. For example, to randomly assign 5 consumers to treatment 1, we would use Table B to select 5 numbers from 01 through 30, and label those subjects as treatment 1. We would then repeat this process for each of the six treatments. An example of this would be: Assign = 5 consumers to each of the six treatments. Label the subjects from 01 through 30. Randomly select 5 numbers for treatment 1, then 5 of the remaining consumers for treatment 2, and so on. Using Table B at line 133, the consumers assigned to treatment 1 are those numbered 4, 5, 7, 1, and 8.

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a) Using the Law of Cosines for this triangle, we can write the equation:
d² = (36)² + (3s)² - 2(36)(3s)cos(60°)

b) The second ship must travel at approximately 4.24 miles per hour to be 42 miles apart from the first ship at noon.

(a) To write an equation that relates the speed s and the distance d between the two ships at noon using the Law of Cosines, we first need to determine the distance each ship has traveled by noon. Since they leave at 9 a.m. and we're interested in the distance at noon, they travel for 3 hours.

Ship 1:
Speed: 12 miles per hour
Distance traveled: 12 miles/hour * 3 hours = 36 miles

Ship 2:
Speed: s miles per hour
Distance traveled: s miles/hour * 3 hours = 3s miles

Now, we can form a triangle where Ship 1 travels 36 miles, Ship 2 travels 3s miles, and the distance between them (d) is the third side. The angle between Ship 1 and Ship 2 is 180° - (53° + 67°) = 60°.

Using the Law of Cosines for this triangle, we can write the equation:
d² = (36)² + (3s)² - 2(36)(3s)cos(60°)

(b) To find the speed s that the second ship must travel so that the ships are 42 miles apart at noon, we can plug d = 42 into our equation from part (a) and solve for s.

42² = (36)² + (3s)² - 2(36)(3s)cos(60°)

Solving for s, we get:
s ≈ 4.24 miles per hour (rounded to two decimal places)

So, the second ship must travel at approximately 4.24 miles per hour to be 42 miles apart from the first ship at noon.

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The conversions are :

a) 2.8 km =  2800 m.

b) 27.55 dm =   2.755 m.

c) 27.9 hm =   2790 m.

d) 275 dam =   2750 m.

By multiplying the value by 1000 will help us to change kilometers (km) to meters (m). In order to change decimeters into meters, it is necessary to divide the figure by 10.

To convert, Note that:

a) 2.8 km to m:

= 2.8 x 1000 m

= 2800 m

b) 27.55 dm to m:

= 27.55 ÷ 10 m

= 2.755 m

c) 27.9 hm to m:

= 27.9 x 100 m

= 2790 m

d) 275 dam to m:

= 275 x 10 m

= 2750 m

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Convert the following to meter

2,8km

27,55dm

27,9hm

275dam

If you pick 3 fruits at random, the mean weight will be greater than 757 grams approximately 1.26% of the time.


To solve this problem, we need to use the properties of the normal distribution. We know that the weights of the fruit are normally distributed with a mean of 753 grams and a standard deviation of 9 grams.
The mean of the sample of 3 fruits will also be normally distributed with a mean of 753 grams and a standard deviation of 3 grams (since we are dividing by the square root of the sample size).

To find the probability that the mean weight of the sample of 3 fruits will be greater than a certain amount, we need to convert this amount to a z-score using the formula:
z = (x - μ) / (σ / √n)

where x is the amount we are interested in, μ is the mean, σ is the standard deviation, and n is the sample size (in this case, 3).

In this case, we want to find the z-score for a mean weight of 757 grams:

[tex]z = \frac{(757 - 753)}{\frac{9}{\sqrt{3}} }} =1.26[/tex]

We can use a standard normal distribution table or calculator to find that the probability of getting a z-score greater than 1.26 is approximately 0.0985, or 9.85%. However, since we are interested in the probability that the mean weight will be greater than 757 grams (not just greater than the mean), we need to add half of the probability of getting exactly 757 grams (which is the mode of the distribution) to this value.
Since the normal distribution is symmetrical, the probability of getting exactly 757 grams is the same as the probability of getting exactly 749 grams (which is the mean minus one standard deviation). Using the same formula as before, we can find the z-score for a weight of 749 grams:

z =  \frac{(749 - 753)}{\frac{9}{\sqrt{3}} }} =-1.26[/tex]

The probability of getting a z-score less than -1.26 is also approximately 0.0985, so the probability of getting exactly 757 grams is approximately 0.197. Half of this value is 0.0985, which we add to the probability of getting a z-score greater than 1.26 to get the final answer:
0.0985 + 0.0985 = 0.197
So the probability of getting a mean weight greater than 757 grams is approximately 0.0985 + 0.197 = 0.2965, or 29.65%.

To convert this probability to weight, we can use a standard normal distribution table or calculator to find the z-score corresponding to a probability of 29.65%. This is approximately 0.56. Using the same formula as before, we can solve for x:

\frac{(x - 753)}{\frac{9}{\sqrt{3}} }} =0.56[/tex]

x ≈ 757.23

So if you pick 3 fruits at random, their mean weight will be greater than 757 grams approximately 1.26% of the time.

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Answer:

I think its 42

Step-by-step explanation:

Hope u get the right answer!

The reflection from green triangle to blue triangle is a reflection over the x-axis

Reflecting a two-dimensional shape over the x-axis is a geometric transformation that represents an image flipping or mirroring itself across the fixed x-axis.

This axis appears as the horizontal marker in a Cartesian coordinate system, providing the reference line to then split the plane into its top and bottom elements.

When which this action is fullfilled, all the y-coordinates of each point within the figure will be reversed while the x-coordinate remains unchanged;

The image of the reflection is attached and the reflection line labeled

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The point V(3, 5) is reflected over the x axis, the new point is V'(3, -5)

Reflection is a type of transformation. Transformation is the movement of a point either up, left, right or down in the coordinate plane.

Reflection is a rigid transformation because it conserves the size and shape of the figure.

If a point A(x, y) is reflected over the x axis, the new point is A'(x, -y)

The coordinate of point V is (3, 5). If the point is reflected over the x axis, the new point is V'(3, -5)

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The equation of circle P include the following: D. x² + (y - 3)² = 4

In Mathematics and Geometry, the standard form of the equation of a circle is represented by the following mathematical equation;

(x - h)² + (y - k)² = r²

Where:

By critically observing the graph of this circle, we have the following parameters:

By substituting the given parameters into the equation of a circle formula, we have the following;

(x - h)² + (y - k)² = r²

(x - 0)² + (y - 3)² = 2²

x² + (y - 3)² = 4.

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The proportion of shoppers who can identify the highly advertised trademark is significantly higher than 40% at the 0.01 level of significance.

Let's break it down step-by-step using the provided information:

1. Hypotheses: - Null hypothesis (H0): p = 0.40 (40% of all shoppers can identify the trademark) - Alternative hypothesis (Ha): p > 0.40 (more than 40% of all shoppers can identify the trademark)

2. Level of significance: - α = 0.01

3. Sample information: - n (sample size) = 18 - x (number of successful identifications) = 13 - p-hat (sample proportion) = x / n = 13 / 18 = 0.7222 4. Test statistic

calculation: - We'll use a one-sample z-test for proportions. - z = (p-hat - p) / sqrt((p * (1 - p)) / n) - z = (0.7222 - 0.40) / sqrt((0.40 * (1 - 0.40)) / 18) - z ≈ 2.88 5. Decision: - Since α = 0.01, we'll compare our test statistic to the critical value from the z-table, which is 2.33 for a one-tailed test. - Our test statistic, z ≈ 2.88, is greater than the critical value of 2.33.

Conclusion: Since our test statistic is greater than the critical value at the 0.01 level of significance, we have enough evidence to reject the null hypothesis (H0). This means that there is sufficient evidence to support the claim that more than 40% of all shoppers can identify a highly advertised trademark.

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