let pq be a focal chord of the parabola 2 x py 4 . show that the circle with diameter pq is tangent to the directrix of the parabola

Answer:

-x(x-1)(x-3)

Step-by-step explanation:

First rearrange the terms:

-x³+4x²-3x

Now we can factor out -x, the common factor

-x(x²-4x+3)

This equals -x(x-1)(x-3)

To solve this problem, we'll use the concept of joint variation and the given information to find the resistance of the refrigerator.

Let's denote:

- Power of the microwave oven as P1 = 600 watts

- Current drawn by the microwave oven as I1 = 5.0 amperes

- Resistance of the microwave oven as R1 = 24 ohms

According to the problem, the power varies jointly as the resistance and the square of the current. Mathematically, this can be represented as:

P1 = k * R1 * I1^2

where k is the constant of variation.

We can solve for k by rearranging the equation:

k = P1 / (R1 * I1^2)

Now, we'll use this value of k to find the resistance of the refrigerator:

Power of the refrigerator, P2 = 200 watts

Current drawn by the refrigerator, I2 = 1.7 amperes

Using the joint variation equation:

P2 = k * R2 * I2^2

Rearranging the equation, we get:

R2 = P2 / (k * I2^2)

Now we can substitute the known values into the equation to find the resistance of the refrigerator:

R2 = 200 / (k * 1.7^2)

First, let's calculate the value of k:

k = 600 / (24 * 5.0^2)

k ≈ 0.5

Now substitute this value into the equation for R2:

R2 = 200 / (0.5 * 1.7^2)

R2 ≈ 200 / (0.5 * 2.89)

R2 ≈ 200 / 1.445

R2 ≈ 138.31

Rounding to the nearest tenth, the resistance of the refrigerator is approximately 138.3 ohms.[tex][/tex]

Hypothesis testing is the smaller the p-value, the more evidence the data provide against the null hypothesis and in favor of the alternative hypothesis.

In hypothesis testing, the p-value represents the probability of obtaining the observed data (or more extreme) if the null hypothesis is true. A smaller p-value indicates that the observed data is less likely to occur by chance under the assumption of the null hypothesis. Therefore, a smaller p-value provides stronger evidence against the null hypothesis and supports the alternative hypothesis, suggesting that there is a significant difference or relationship in the data.

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Answer:

i -2

ii 2

Step-by-step explanation:

y=mx+c

y=-2x+2

m=-2

y-intercept

let x=0

y=-2(0)+2

y=2

Answer:

gradient = - 2 , y- intercept = 2

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope (gradient ) and c the y- intercept )

y = - 2x + 2 ← is in slope- intercept form

with gradient m = - 2 and y- intercept c = 2

A unit vector in the direction of the given vector [-6, 4, -4] is:

[(-6/√68), (4/√68), (-4/√68)].

A unit vector is a vector that has a magnitude of 1 and is used to represent direction in a specific coordinate system.

Find the magnitude of the vector [-6, 4, -4]. The magnitude formula is √(x^2 + y^2 + z^2).
Magnitude = √((-6)^2 + (4)^2 + (-4)^2) = √(36 + 16 + 16) = √(68)

Divide each component of the vector by its magnitude to find the unit vector.
Unit vector = [(-6/√68), (4/√68), (-4/√68)]

Therefore, a unit vector in the direction of the given vector [-6, 4, -4] is [(-6/√68), (4/√68), (-4/√68)].

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The most appropriate method to test the hypothesis that, on average, students spend more time studying for SCMA301 than SCMA212 is the T-test for paired sampling.

The T-test for paired sampling is suitable for comparing the means of two related samples. In this case, the same group of 30 students is being observed and compared in two different situations: studying for SCMA212 and studying for SCMA301.

By using the paired sampling T-test, we can analyze the differences between the two sets of paired data (hours spent studying) for each student.

This method accounts for individual differences among students, such as learning styles or study habits, by comparing each student's performance in both courses. It allows us to determine whether there is a significant difference in the mean study hours between SCMA212 and SCMA301 for the same group of students. If the T-test reveals a statistically significant difference, it supports the hypothesis that, on average, students spend more time studying for SCMA301 than SCMA212.

Other methods mentioned, such as the Test on two proportions, Test on the linear regression slope, and Chi-square test for independence between two categorical variables, are not appropriate for comparing continuous numerical data, as in this case.

The T-test assuming unequal variances for two independent samples is also not ideal because we are comparing the same group of students in two different courses.

Therefore, to determine if students spend more time studying for SCMA301 than SCMA212, the most suitable approach is to use a paired t-test to compare the average study times of students in both courses.

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Complete question here:

30 students recorded the number of hours they spent studying during the semester outside of class for SCMA212. The next semester The 30 students took SCMA301and the recorded the no of hours spent studying outside of class for it. I hypothesize that on the same 30 students took SCMA301 and recorded the number the average students spend more time studying for SCMA301 than SCMA212 The most appropriate method to test the above hypothesis is?

T-test for paired sampling

Test on two proportions

Test on the linear regression slope

Chi-square test for independence between two categorical variables

T-test assuming unequal variances for two independent samples

The system of equations are solved

Given data ,

Let the equation be represented as A

Now , the value of A is

a)

9m + 7 = 8m

On simplifying the equation , we get

Subtracting 7 on both sides , we get

9m = 8m - 7

Subtracting 8m on both sides , we get

m = -7

b)

( 1/3 ) + ( 5/3 )r - r = 2r

On simplifying the equation , we get

Adding r on both sides , we get

( 1/3 ) + ( 5/3 )r = 3r

Subtracting ( 5/3 )r on both sides , we get

( 1/3 ) = ( 4r / 3 )

Multiply by 3 on both sides , we get

4r = 1

Divide by 4 on both sides , we get

r = 1/4

c)

1 = 2k - 1

On simplifying the equation , we get

Adding 1 on both sides , we get

2k = 2

Divide by 2 on both sides , we get

k = 1

Hence , the equations are solved

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The algorithm ensures that g is a generator, satisfying the necessary conditions.

To show that the algorithm produces a generator, we need to verify two conditions:

The algorithm selects a number, g, that is coprime to p-1.

The algorithm checks if g raised to the power of (p-1)/q is not congruent to 1 modulo p for each prime factor, q, of p-1.

Let's go through the steps of the algorithm to demonstrate these conditions:

Start with the prime factorization of p-1: p-1 = q1^a1 * q2^a2 * ... * qn^an, where q1, q2, ..., qn are distinct prime factors.

For each prime factor, q, calculate g = h^((p-1)/q) modulo p, where h is a randomly selected number between 2 and p-1.

Check if g is congruent to 1 modulo p. If it is, go back to step 2 and select a different h.

Repeat steps 2 and 3 until g is not congruent to 1 modulo p for each prime factor, q.

Now let's analyze these conditions:

By raising h to the power of (p-1)/q for each prime factor, the resulting g is guaranteed to be coprime to p-1. This is because g is not divisible by any prime factor of p-1.

For each prime factor, q, g^(p-1)/q is not congruent to 1 modulo p. This is because g^(p-1)/q is congruent to h^((p-1)/q * (p-1)/q) modulo p, and since h^((p-1)/q) is coprime to p, raising it to the power of (p-1)/q * (p-1)/q will not result in 1 modulo p.

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The Taylor polynomial T₅(x) is 0.99500416 and by use the Error Bound the maximum possible size of the error is 49943.1 × 10⁻⁷.

What is Taylor Series?

The Taylor series or Taylor expansion of a function in mathematics is the infinite sum of terms represented in terms of the function's derivatives at one particular point. The function and the sum of its Taylor series are roughly equivalent for the majority of typical functions at this point.

Taylor series or Taylor expansion:

Infinity ∑ (n = 0) fⁿ(a)/n! (x - a)ⁿ

Where,

n! = factorial of n

a = real or complex number

fⁿ(a) = nth derivative of function f evaluated at the point a.

As given function is,

f(x) = cosx, a = 0, x = 0.1

Taylor polynomial of degree 'n' for f(x) center a,

Tₙ(x) = f(a) + f'(a)(x - a) + f''(a)/2 (x - a)² + f'''(a)/3 (x - a)³ + ......+ fⁿ⁻¹(a)/(n - 1)! (x - a)ⁿ⁻¹ + fⁿ(a)/n! (x - a)ⁿ

Evaluate values as follows:

f(x) = cosx ⇒ f(0) = 1

f'(x) = -sinx ⇒ f'(0) = 0

f''(x) = -cosx ⇒ f''(0) = -1

f'''(x) = sinx ⇒ f'''(0) = 0

f⁴(x) = cosx ⇒ f⁴(0) = 1

f⁵(x) = -sinx ⇒ f⁵(0) = 0

Substitute obtained values in Taylor series,

T₅(x) = 1 + (0) (x - 0) + (-1)/2 (x - 0)² + 0 + 1/24(x - 0)⁴ + 0

T₅(x) = 1 -1/2x² + 1/24x⁴

At x = 0.1

T₅(0.1) = 1 -1/2(0.1)² + 1/24(0.1)⁴

T₅(0.1) = 1 - 0.005 + 4.16 × 10⁻⁶

T₅(0.1) = 0.99500416

Hence, the Taylor polynomial T₅(x) is 0.99500416.

Evaluate the maximum possible size of the error:

cos(0.1) = 0.99999847

T₅(0.1) = 0.99500416

Icos(0.1) - T₅(0.1)I = 0.99999847 - 0.99500416

Icos(0.1) - T₅(0.1)I = 0.00499431

Icos(0.1) - T₅(0.1)I = 49943.1 × 10⁻⁷.

Hence, the Error Bound the maximum possible size of the error is 49943.1 × 10⁻⁷.

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The correct slope of the tangent line to y = f(t)g(2) at the point where r = 3 is 21, not 14 as Adam entered.

To find the slope of the tangent line to the function y = f(t)g(2) at the point where r = 3, we can use the product rule of differentiation. Let's analyze Adam's approach and identify the mistake(s).

Adam's mistake is in assuming that the slope of the tangent line to y = f(x)g(x) at the point where x = 3 is simply the product of the slopes of the individual tangent lines to f(x) and g(x) at x = 3. This assumption is incorrect because the product rule accounts for the interaction between the two functions.

To find the slope of the tangent line to y = f(t)g(2) at the point where r = 3, we need to apply the product rule:

(dy/dt) = (f'(t) * g(2)) + (f(t) * g'(2))

Given the information provided, we know:

f(3) = 5

f'(3) = 2

g(3) = -7

g'(3) = 7

Now, let's substitute these values into the product rule equation:

(dy/dt) = (f'(t) * g(2)) + (f(t) * g'(2))

(dy/dt) = (2 * g(2)) + (f(t) * 7)

(dy/dt) = (2 * g(2)) + (5 * 7)

(dy/dt) = (2 * g(2)) + 35

Since we are interested in the slope at the point where r = 3, we substitute r = 3 into the equation:

(dy/dt) = (2 * g(2)) + 35

(dy/dt) = (2 * (-7)) + 35

(dy/dt) = -14 + 35

(dy/dt) = 21

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The standard deviation of the sampling distribution of the difference in the two sample proportions is 0.034.

The standard deviation of the sampling distribution of the difference in two sample proportions can be calculated using the following formula:

σ= √[(P₁ (1 - P₁) / n₁) + (P₂ (1 - P₂) / n₂)]

Where:

P₁ and P₂ are the sample proportions of US and Spanish teens with a TV in their bedrooms, respectively.

n₁ and n₂ are the sample sizes of the US and Spanish teens, respectively.

In this case:

P₁ = 0.77 (77% of US teens have a TV in their bedrooms)

P₂ = 0.59 (59% of Spanish teens have a TV in their bedrooms)

n₁ = 500 (sample size of US teens)

n₂ = 300 (sample size of Spanish teens)

Plugging in these values into the formula:

σ = √[(0.77× (1 - 0.77) / 500) + (0.59 × (1 - 0.59) / 300)]

Calculating the expression inside the square root:

σ= √[(0.77 × 0.23 / 500) + (0.59 × 0.41 / 300)]

σ = 0.034

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The shaded region is more than 3 m from P and closer to Q than to P.

As see in the image:

Point Q is 3 m far form the shaded region.

The shaded region more than 3 m from P and closer to Q than to P.

Therefore, the  shaded region is close to Q.

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The equation of the tangent line to the curve y = ln(x^2 - 5x + 1) at the point (5,0) is y = 0.

To find the equation of the tangent line to the curve y = ln(x^2 - 5x + 1) at the point (5,0), we need to find the slope of the tangent line at that point.

The derivative of y = ln(x^2 - 5x + 1) is:

y' = (2x - 5)/(x^2 - 5x + 1)

At the point (5,0), we have:

y' = (2(5) - 5)/(5^2 - 5(5) + 1) = 0

So the slope of the tangent line at (5,0) is 0.

The equation of the tangent line can be written as:

y - y1 = m(x - x1)

where (x1, y1) is the point of tangency and m is the slope.

Since the slope is 0, we have:

y - 0 = 0(x - 5)

which simplifies to:

y = 0

Therefore, the equation of the tangent line to the curve y = ln(x^2 - 5x + 1) at the point (5,0) is y = 0.

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Jean is currently 24 years old, while Dean is 38 years old.

Let's assume Dean's current age is D. According to the given information, Jean's age is 2 less than half of Dean's age, which can be represented as (1/2)D - 2.

In two years, Jean will be half of Dean's age. So, in two years, Jean's age will be (1/2)(D + 2), and Dean's age will be D + 2.

Based on the given information, we can set up the equation:

Jean's age + Dean's age = 62

Substituting the expressions for their current ages, we get:

(1/2)D - 2 + D = 62

Multiplying the equation by 2 to eliminate the fraction, we have:

D - 4 + 2D = 124

Combining like terms, we get: 3D - 4 = 124

Adding 4 to both sides, we have: 3D = 128

Dividing both sides by 3, we find: D = 42.67

Since ages cannot be fractional, we can conclude that Dean's current age is 38 (the closest whole number), and Jean's age is (1/2)(38) - 2 = 24.

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The probability of selecting a card that is less than 7 or a spade is 4/13.

There are 52 cards in a deck, of which 13 are spades and 4 are less than 7 in each suit (2, 3, 4, 5, 6). There are 3 suits that have cards less than 7, so there are a total of 12 cards less than 7 in the deck.

To find the probability of selecting a card that is less than 7 or a spade, we need to add the probabilities of these two mutually exclusive events.

The probability of selecting a card less than 7 is 12/52, because there are 12 cards less than 7 in the deck.

The probability of selecting a spade is 13/52, because there are 13 spades in the deck.

Since we only want to count the spades that are not also less than 7, we need to subtract the probability of selecting a spade that is also less than 7. There are 9 cards that are both spades and less than 7 (2, 3, 4, 5, 6 of spades), so the probability of selecting a spade that is also less than 7 is 9/52.

Therefore, the probability of selecting a card that is less than 7 or a spade is:

P(less than 7 or spade) = P(less than 7) + P(spade) - P(less than 7 and spade)

= 12/52 + 13/52 - 9/52

= 16/52

= 4/13

So the probability of selecting a card that is less than 7 or a spade is 4/13.

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These tests have alternative hypotheses that explicitly state X is less than the specified value.

The alternative hypothesis (Ha) is the other answer to your research question. It claims that there's an effect in the population. Often, your alternative hypothesis is the same as your research hypothesis. In other words, it's the claim that you expect or hope will be true

Left-tailed tests are hypothesis tests where the alternative hypothesis suggests that the population parameter is less than a specified value. Therefore, the correct left-tailed tests among the answer choices are:

B) H0: X = 19.7, Ha: X < 19.7

C) H0: X = 11.2, Ha: X < 11.2

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the probability of drawing a queen or a spade from a standard deck of 52 cards is 7/26.

There are four queens and 13 spades in a standard deck of 52 cards. However, one of the queens is a spade, so we need to subtract this card from the count to avoid double-counting it.

Therefore, there are 4 - 1 = 3 queens that are not spades and 13 - 1 = 12 non-queen spades.

The probability of drawing a queen or a spade can be calculated by adding the probability of drawing a queen to the probability of drawing a spade and subtracting the probability of drawing the queen of spades twice (since it belongs to both groups):

P(queen or spade) = P(queen) + P(spade) - P(queen of spades)

= 3/52 + 12/52 - 1/52

= 14/52

= 7/26

Therefore, the probability of drawing a queen or a spade from a standard deck of 52 cards is 7/26.

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The resulting value gives us the value of h at y = -8. In this case, plunging -8 into the expression gives us 253.
The given function is h(y) = 4y^2 - 3. To evaluate h(-8), we substitute y = -8 into the function:

h(-8) = 4(-8)^2 - 3

     = 4(64) - 3

     = 256 - 3

     = 253

Therefore, h(-8) = 253.

We simply substitute -8 into the expression for y in the function and then evaluate the resulting expression. The resulting value gives us the value of h at y = -8. In this case, plugging -8 into the expression gives us 253.
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The sample that would be the most appropriate for the survey is : Every 5th student in the 7th grade.

Therefore option C is correct.

we have the following:

Becker's data: 5, 2, 4, 1, 1, 4, 5, 3, 2, 1

We then find the mean by summing  up all the data points and then divide by the total number of data points:

Mean for Becker = (5 + 2 + 4 + 1 + 1 + 4 + 5 + 3 + 2 + 1) / 10

mean for Becker = 28 / 10

Mean for Becker= 2.8

We also have that Kayla's data: 2, 3, 1, 1, 4, 1, 3, 5, 5

The mean for Kayla = (2 + 3 + 1 + 1 + 4 + 1 + 3 + 5 + 5) / 9

The mean for Kayla = 25 / 9

The mean for Kayla  = 2.78

If we compare  the both means, we notice  that the mean number of games played per day for both Becker and Kayla is estimated to be  2.8.

In conclusion, Option C ensures a random selection of students from all the classes in the 7th grade.

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The value of ∫[6, -10]g(x)dx is 2.

We are given two integrals: ∫[-104, -10]g(x)dx = -3 and ∫[6, 4]g(x)dx = 5.

We need to find the value of ∫[6, -10]g(x)dx.

Notice that the desired integral's interval is just the union of the given intervals.

Therefore, we can obtain the value of the desired integral by summing the values of the two given integrals: ∫[6, -10]g(x)dx = ∫[-104, -10]g(x)dx + ∫[6, 4]g(x)dx = -3 + 5 = 2.

Thus, the value of ∫[6, -10]g(x)dx is 2.

The interval of an integral refers to the range over which the integration is performed.

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There are 2 × 3^n-1 - 2^n-1 n-digit ternary sequences in which at least one pair of consecutive digits are the same.

The number of n-digit ternary sequences in which at least one pair of consecutive digits are the same can be found using the principle of inclusion-exclusion. We first calculate the total number of n-digit ternary sequences, which is 3ⁿ since each digit can take on three possible values (0, 1, or 2). Next, we calculate the number of n-digit ternary sequences in which no pair of consecutive digits are the same.

To do this, we observe that for any such sequence, the first digit can be any of the three possible values. However, each subsequent digit can only be one of the two values that are different from the previous digit. Therefore, there are 3 choices for the first digit and 2 choices for each subsequent digit, resulting in a total of 3 × 2^n-1 sequences with no consecutive equal digits.

Now, we need to subtract the number of sequences with no consecutive equal digits from the total number of sequences. However, this will also subtract the number of sequences in which two pairs of consecutive digits are the same (e.g., 1101). Therefore, we need to add back in the number of sequences with two pairs of consecutive digits that are the same.

To do this, we observe that there are 2^n-2 ways to choose the positions of the two pairs of consecutive equal digits, and there are 2 choices for the digits in each pair (since they must be the same). Therefore, there are a total of 2^n-1 sequences with two pairs of consecutive equal digits.

Using the principle of inclusion-exclusion, the number of n-digit ternary sequences in which at least one pair of consecutive digits are the same is:

3ⁿ - 3 × 2^n-1 + 2^n-1

This can be simplified to:

2 × 3^n-1 - 2^n-1

Therefore, there are 2 × 3^n-1 - 2^n-1 n-digit ternary sequences in which at least one pair of consecutive digits are the same.

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The joint density of the polar coordinates R and Θ is f(R,Θ) = 1/(πR), where 0 ≤ R ≤ 1 and 0 ≤ Θ ≤ 2π.

To find the joint density of the polar coordinates R and Θ, we need to use the transformation of variables formula for joint densities. Let us define the transformation from Cartesian coordinates (X, Y) to polar coordinates (R, Θ) as follows:

R = (X^2 + Y^2)^(1/2)

Θ = tan^-1(Y/X)

We can solve for X and Y in terms of R and Θ as follows:

X = R cos(Θ)

Y = R sin(Θ)

We can then compute the Jacobian of this transformation:

|dX/dR dX/dΘ| |cos(Θ) -R sin(Θ)|

| | = | |

|dY/dR dY/dΘ| |sin(Θ) R cos(Θ)|

The determinant of this matrix is R, so the joint density of R and Θ can be obtained as follows:

f(R,Θ) = f(X,Y) * |Jacobian|

= (1/π) * (1/(X^2 + Y^2)) * R

= (1/π) * (1/R^2) * R

= 1/(πR), 0 ≤ R ≤ 1, 0 ≤ Θ ≤ 2π

Therefore, the joint density of the polar coordinates R and Θ is f(R,Θ) = 1/(πR), where 0 ≤ R ≤ 1 and 0 ≤ Θ ≤ 2π.

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Therefore, it is important to carefully consider the significance level and the power of the test to minimize the risk of Type I errors.

If the conclusion from the hypothesis test is to reject the null hypothesis, which means that there is evidence to suggest that the mean running time has increased, but in reality it has not increased, then the conclusion is a Type I error. This is because a Type I error occurs when the null hypothesis is incorrectly rejected, leading to a false positive result. In this case, the manufacturer may make changes to the production process based on the false belief that the mean running time has increased, leading to potential losses in time, money, and resources. Therefore, it is important to carefully consider the significance level and the power of the test to minimize the risk of Type I errors.

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The correct statement regarding the y-intercepts of the functions is given as follows:

C. The y-intercept of item II is less than the y-intercept of item I.

The y-intercept of a function is the numeric value of a function when the input assumes a value of zero.

For item I, the y-intercept is given as follows:

y = 5, as when x = 0, y = 5.

For item II, the y-intercept is given as follows:

y = 0, as the graph crosses the y-axis at y = 0.

As 5 > 0, we have that the correct statement is given by option C.

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To find the acceptance set A for a significance level of α = 0.09 and rejection set R of the form { |K - E(K)| > c}, we first need to calculate the expected value and variance of K.

(A) Since the coin is fair, E(K) = 50 and Var(K) = 25/2. Using the Central Limit Theorem, we can approximate K as a normal distribution with a mean of 50 and a standard deviation of 2.5. We can then find the value of c such that P(|K - 50| > c) = 0.09/2 = 0.045. Solving for c, we get c = 3.325. Therefore, the acceptance set A is {45 < K < 55}.

(b) For a significance level of α = 0.018 and rejection set R of the form {K > λ}, we again use the Central Limit Theorem to approximate K as a normal distribution with a mean of 50 and a standard deviation of 2.5. We can then find the value of λ such that P(K > λ) = 0.018. Using a normal distribution table or calculator, we find λ to be approximately 60.5. Therefore, the acceptance set A is {K ≤ 60}.

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Answer:

Nevaeh was 5 mins in the bank.

Step-by-step explanation:

straight line mean is 'stop'.

The volume of the cube is approximately 11.44 cubic centimeters.

To calculate the volume of a cube, we need to raise the length of one of its edges to the power of 3. In this case, the length of the edge is given as 2.23 cm.

The volume of the cube can be calculated as follows:

Volume = (Edge length)^3

Volume = (2.23 cm)^3

Volume = 2.23 cm * 2.23 cm * 2.23 cm

Volume ≈ 11.44 cm³

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To compare the posttreatment t-test scores of two groups of subjects who received different treatments, the appropriate analytic statistic would be a two-sample t-test. An independent samples t-test would be appropriate for analyzing the data.

This test is used to determine whether there is a statistically significant difference between the means of two independent groups. The two-sample t-test assumes that the samples are independent, normally distributed, and have equal variances.

If the probability is lower than a predetermined significance level (usually 0.05), then we reject the null hypothesis that the two group means are equal and conclude that there is a statistically significant difference between the treatments.

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X follows a continuous uniform distribution from 1 to 5. Then, the conditional probability of x being greater than 2.5 given that it is less than or equal to 4 is 0.5.

We can use Bayes' theorem and the definition of conditional probability to solve this problem.

Let A be the event that x is less than or equal to 4, and B be the event that x is greater than 2.5. Then, we want to find P(B|A), the conditional probability of B given A.

By Bayes' theorem, we have:

P(B|A) = P(A|B) * P(B) / P(A)

We can find each of these probabilities separately:

P(A) is the probability that x is less than or equal to 4, which is given by the cumulative distribution function(CDF) of the uniform distribution:

P(A) = (4 - 1) / (5 - 1) = 0.75

P(B) is the probability that x is greater than 2.5, which is also given by the CDF of the uniform distribution:

P(B) = (5 - 2.5) / (5 - 1) = 0.375

For P(A|B), we can use the formula for conditional probability:

P(A|B) = P(A and B) / P(B)

The probability of A and B can be found using the joint probability density function (PDF) of the uniform distribution:

f(x) = 1 / (5 - 1) = 0.25, for 1 <= x <= 5

Then, we have:

P(A and B) = ∫2.5^4 f(x) dx = ∫2.5^4 0.25 dx = 0.375

Therefore, we have:

P(A|B) = P(A and B) / P(B) = 0.375 / 0.375 = 1

Finally, we can substitute these values into Bayes' theorem to find P(B|A):

P(B|A) = P(A|B) * P(B) / P(A) = 1 * 0.375 / 0.75 = 0.5

Therefore, the conditional probability of x being greater than 2.5 given that it is less than or equal to 4 is 0.5.

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