Charlie's teacher claims that he does not study and just guesses on exams. on an exam with 201 true-false questions, Charlie answered 53.7% of the questions correctly. Calculations using these results show that if he were really just guessing, there would be roughly one chance and seven that he would do this well. Is there a statistically significant evidence against the teachers claim that Charlie is just guessing? Why or why not?

Answer:

Step-by-step explanation:

True.

A Data Flow Diagram (DFD) is a graphical representation of a system or a process that shows how data flows through different components of the system. It provides a visual representation of the system's inputs, processes, and outputs.

While a DFD can provide information on the data that flows between the different components of a system, it does not provide any information about the timing or sequence of those processes. A DFD is a static diagram that only depicts the flow of data in a system and does not provide any indication of when or how frequently processes occur

The value of p(2) = 7, p(0) = -3 , p(-1) = -17 and p(-2) = -53 when polynomial is p(x) = 2x³ - 5x² + 7x - 3 with one variable.

Given that,

The polynomial is p(x) = 2x³ - 5x² + 7x - 3

We have to find the value of p(2), p(0), p(-1) and p(-2).

We know that,

Take polynomial,

p(x) = 2x³ - 5x² + 7x - 3

Now, to find p(2) take x = 2 in polynomial

By substituting,

p(2) = 2(2)³ - 5(2)² + 7(2) - 3

p(2) = 16 - 20 + 14 - 3

p(2) = 30 - 23

p(2) = 7

Now, to find p(0) take x = 0 in polynomial

p(0) = 2(0)³ - 5(0)² + 7(0) - 3     [multiplication]

p(0) = 0 - 0 + 0 - 3

p(0) = -3

Now, to find p(-1) take x = -1 in polynomial

p(-1) = 2(-1)³ - 5(-1)² + 7(-1) - 3

p(-1) = -2 - 5 - 7 - 3                  [subtraction]

p(-1) = -17

Now, to find p(-2) take x = -2 in polynomial

p(-2) = 2(-2)³ - 5(-2)² + 7(-2) - 3

p(-2) = -16 - 20 - 14 - 3

p(-2) = -53

Therefore, The value of p(2) = 7, p(0) = -3 , p(-1) = -17 and p(-2) = -53.

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There are different ways to approach this question, but one possible method is to conduct a one-sample t-test using the formula t = (M - μ) / (s / sqrt(n)), where M is the sample mean, μ is the population mean (null hypothesis), s is the sample standard deviation, and n is the sample size. With the given information, the t-value is (1.6 - 0) / (0.8 / sqrt(16)) = 8.

The degrees of freedom is n - 1 = 15. Using a t-table or a calculator, the p-value for a one-tailed test with 15 degrees of freedom and a t-value of 8 is much smaller than .05 (e.g., < .0005). Therefore, we can reject the null hypothesis and conclude that women find a sense of humor to be important in a long-term relationship at a .05 level of significance. The correct answer is b.

Test 2 is associated with smaller critical values. The critical values for a t-test depend on the level of significance, the degrees of freedom, and whether the test is one-tailed or two-tailed. In general, a smaller level of significance or a larger sample size leads to smaller critical values (i.e., it becomes easier to reject the null hypothesis). Since Test 2 has a larger sample size than Test 1, it is more likely to detect a significant effect (i.e., have a smaller Type II error rate), and therefore requires smaller critical values to achieve a .05 level of significance. The correct answer is b.

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Answer: mean= 24.2 and the standard deviation is 6.0 :)

Step-by-step explanation: this is because when you solve for the equation the result for the standard deviation is 6.3 but if you round the number, it becomes 6.0, being the standard deviation.

The expected count of women who ranked the dark-skinned model as the most beautiful is 100.

Since there are three models, we would expect 100/3 = 33.33% of the women to have ranked each model as the most beautiful if there were no relationship between skin color and beauty judgements. Thus, we can calculate the expected count of women who ranked the dark-skinned model as the most beautiful as follows:

Expected count = 300 * 33.33% = 100

Therefore, we can expect that 100 women out of the 300 sampled would have ranked the dark-skinned model as the most beautiful if there were no relationship between skin color and beauty judgements. The observed count of 55 women who ranked the dark-skinned model as the most beautiful is higher than the expected count, which suggests that there may be a relationship between skin color and beauty judgements. However, further statistical tests would be needed to confirm this.

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Answer:

18.8

Step-by-step explanation:

To find the range of a data set, you subtract the smallest value from the largest value.

In this case, the smallest value is 2.8 and the largest value is 21.6.

Range = largest value - smallest value = 21.6 - 2.8 = 18.8

Therefore, the range of the data set is 18.8.

The system of the equations that describes the situation can be represented as: j + g = 23, j - 2g = - 4

A system of equation is a linear algebraic expression that are represented in variables with their coefficients. A system of equation usually comprises of two equation and these equation are attached to their individual constant.

This implies that, we can have a system of equations where x and y are the variables and the constant can be any number. From the given information, let's assume that the variable for Jodi is "j" and Jerry is "g", then:

j + g = 23

If Jodi buys 4 less than twice number of Jerry's games, then we can have:

j = 2g - 4

j - 2g = - 4

Therefore, the system of the equations can be represented as:

j + g = 23        ------- (1)

j - 2g = - 4     --------- (2)

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The angles in the trapezoid are as follows:

m∠D = 21°

m∠A = 159°

m∠B =  159°

Isosceles trapezoid is a trapezoid in which the top and bottom sides are parallel, while the remaining two non-parallel sides have the same length.

Adjacent angles of the isosceles trapezoid are congruent.

Hence,

m∠D = 21 degrees

Any lower base angles is supplementary to any upper base angles.

m∠A = 180 - 21 = 159 degrees

m∠B = 180 - 21 = 159 degrees

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Answer:

x - 2 > 7

Step-by-step explanation:

Substitute the value 5 for x.

x = 5

-2x ≤ 12

-2(5) ≤ 12

-10 ≤ 12  True

-10 is less than or equal to 12

x - 2 > 7

5 - 2 > 7

3 > 7       Not true

3 is greater than 7

2x < 12

2(5) < 12

10 < 12   True

10 is less than 12

x - 7 < 2

5 - 7 < 2

-2 < 2   True

-2 is less than 2

Answer:

RS = 15 units

Step-by-step explanation:

Given:
RT = 9
TS = 12

To find: Length of RS

Proof:

In right triangle RTS,

[tex]RT^{2} + TS^{2} = RS^{2}[/tex]
[tex]9^{2} + 12^{2} = RS^{2}[/tex]
[tex]81 + 144 = RS^{2}[/tex]
[tex]225 = RS^{2}[/tex]
[tex]\sqrt{225} = RS[/tex]
15 = RS
∴ The length of RS is 15 units.

Answer:

The answer is 460m²

Step-by-step explanation:

Area of rectangle =Length × Width

A=20×23

A=460m²

Answer:460

Step-by-step explanation: 20

                                          x 23

                                          _____

                                               60

                                             +400

A ratio is a relationship in which for every x units of one quantity there are y units of another quantity. Ratios are considered to be equivalent if they express the same x/y. A ratio compares 2 like or unlike quantities by division. Ratios can be shown visually using a graph or a ratio table.

The expression states that for every linear polynomial p with real coefficients and a nonzero coefficient of x, there is exactly one real root r.

For all linear polynomials with real coefficients and a nonzero coefficient of x, there exists exactly one real root. This can be expressed using the universal quantifier "for all" and the existential quantifier "there exists", connected by the logical connective "and". Additionally, the statement "exactly one real root" can be expressed using the quantifier "there exists" and the logical connective "and".

Using quantifiers and logical connectives, we can express the given fact as follows:

∀p ∃!r ((isLinearPolynomial(p) ∧ hasRealCoefficients(p) ∧ coefficientOfX(p) ≠ 0) → hasRealRoot(p, r))

Explanation:

- ∀p: For every polynomial p
- ∃!r: There exists exactly one real root r
- isLinearPolynomial(p): p is a linear polynomial (degree 1)
- hasRealCoefficients(p): p has real coefficients
- coefficientOfX(p) ≠ 0: The coefficient of x in p is nonzero
- hasRealRoot(p, r): p has a real root r

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The geometric mean is used to determine the average annual percent increase, as it accurately accounts for compounding growth over multiple periods. The correct option is E.

The measure of central tendency that is typically used to determine the average annual percent increase is the arithmetic mean. This measure takes the sum of all the values and divides it by the total number of values.

It is important to note that other measures, such as the geometric mean, can also be used in certain situations. But for most cases, the arithmetic mean is the go-to measure for determining the average annual percent increase.

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The sketch of the graph of the logarithm function is added s an attachment

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

y = 3log₂(-x) - 9

The above equations is an illustration of a logarithm function that has been transformed using the following

Next, we plot the graph using a graphing tool

The graph of the logarithm function is added as an attachment

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The linear regression equation that best fits the data set is y = 126x + 462.

To find the linear regression equation, we can use the following steps:

Calculate the slope and the y-intercept..

In this case, we have the following values:

y₂ = 4,920

y₁ = 650

x₂= 20

x₁ = 2

Substituting these values into the formula, we get the following:

m = (4,920 - 650)/(20 - 2) = 2,270/18 = 126.11

The y-intercept is calculated using the following formula:

b = y - mx

In this case, we can use the point (2, 650) because it is the first point in the table. Substituting these values into the formula, we get the following:

b = 650 - 126.11(2) = 461.89

The linear regression equation that best fits the data set is y = 126.11x + 461.89. Round each parameter to the nearest whole number, we get y = 126x + 462.

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The number of views of a video on the internet is shown in the table below as a function of the time since the video was posted (in hours).

Number of hours, x

Number of views, y

2

650

5

1,280

8 12

2,140

3,120

17

4,050

20

4,920

Find a linear regression equation, in the form y=ax+b, that best fits this data set. Round each parameter to the nearest whole number.

We should add 25 to make it a perfect square.

Given is an expression 4x²+20x, we need to add a term to make it a perfect square,

So,

4x²+20x

= (2x)² + 2 × 2 × 5 × x

Add and subtract 25.

= (2x)² + 2 × 2 × 5 × x + 25 - 25

= (2x)² + 2 × 2 × 5 × x + 5² - 25

= (2x+5)² - 25

Hence we should add 25 to make it a perfect square.

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The percentage of pregnancies last shorter than 38 weeks is (c) 30.85%.

We can use the normal distribution formula to calculate the percentage of pregnancies that last shorter than 38 weeks:

z = (x - μ) / σ

where z is the z-score, x is the value we are interested in (in this case, x = 38), μ is the mean (μ = 39), and σ is the standard deviation (σ = 2).

Substituting the values, we get:

z = (38 - 39) / 2 = -0.5

We can now use a standard normal distribution table or a calculator to find the percentage of values that fall below the z-score of -0.5. The area to the left of -0.5 is approximately 30.85%.

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the equation of the ellipse in standard form with the given properties is:

x²/16 + y²/16 = 1

What is ellipse?
An ellipse is a closed curve that is symmetric about its center. It is a type of conic section, which is formed by intersecting a cone with a plane. An ellipse is defined as the set of all points in a plane for which the sum of the distances to two fixed points (called the foci) is constant.

To write the equation of an ellipse in standard form with the center at the origin, foci at (-1, 0), and co-vertices at (0, -4), we can use the following information:

Center: (h, k) = (0, 0) (since the center is at the origin)

Foci: (c, 0) = (-1, 0)

Co-vertices: (0, b) = (0, -4)

Let's denote "a" as the distance from the center to the vertices and "c" as the distance from the center to the foci.

From the given information, we can deduce that:

a = 4 (since the distance from the center to the co-vertices is 4)

c = 1 (since the distance from the center to the foci is 1)

The standard form equation for an ellipse with a horizontal major axis and center at the origin is:

x²/a² + y²/b² = 1

Substituting the values we found:

x²/4² + y²/(-4)² = 1

Simplifying the equation:

x²/16 + y²/16 = 1

Thus, the equation of the ellipse in standard form with the given properties is:

x²/16 + y²/16 = 1

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The appropriate inference method to test for a difference between the average IQ scores of identical twins raised by birth parents and in a foster home is paired t-test.

A paired t-test is used when comparing two sets of observations that are paired or matched in some way, such as in the case of identical twins. In this scenario, the twins are matched based on their genetic makeup, and their IQ scores are compared based on their different upbringing environments (birth parents vs. foster home). The paired t-test allows us to analyze the difference between the paired observations (IQ scores) and determine if there is a statistically significant difference between the two groups.

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Tom does have enough fertilizer to cover the triangular area, as the triangular area is of 179.6 m², and he has 300 m² of fertilizer.

We are given the three sides of the triangle, hence the first step in obtaining the area is obtaining the semi-perimeter, which is half the perimeter, hence:

s = (26 + 20 + 18)/2

s = 32 m.

The area of the triangle is then obtained as follows:

[tex]A = \sqrt{s(s - a)(s - b)(s - c)}[/tex]

[tex]A = \sqrt{32(32 - 26)(32 - 20)(32 - 18)}[/tex]

A = 179.6 m².

Hence Tom does have enough fertilizer to cover the triangular area, as the triangular area is of 179.6 m², and he has 300 m² of fertilizer.

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The term (9n+1) grows without bound as n approaches infinity, the limit does not exist. Therefore, the sequence diverges.

To find the limit of the given sequence, we can use the ratio test:

StartFraction
(9(n+1)+1)! / (9(n+1))!
Over
(9n+1)! / (9n)!
EndFraction

Simplifying the expression, we get:

StartFraction
(9n+10)(9n+9)(9n+8)...(9n+2)(9n+1)
Over
(9n+1)(9n)(9n-1)...(2)(1)
EndFraction

The terms cancel out and we are left with:

StartFraction
(9n+10)(9n+9)
Over
9n(9n+1)
EndFraction

Taking the limit as n approaches infinity, we get:

lim (n → ∞) StartFraction
(9n+10)(9n+9)
Over
9n(9n+1)
EndFraction

= lim (n → ∞) StartFraction
81n² + 81n + 90
Over
81n² + 9n
EndFraction

= lim (n → ∞) StartFraction
n² + n + 10 / n² + n / 9
EndFraction

As n approaches infinity, the higher order terms dominate, so we can ignore the constants and simplify the expression to:

lim (n → ∞) StartFraction
n² / n²
EndFraction

= 1

Since the limit exists and is finite, the sequence converges. Therefore, the limit of the sequence is 1.

To find the limit of the given sequence or determine if it diverges, consider the sequence:

a_n = (9n+1)! / (9n)!

We can rewrite the sequence using the properties of factorials:

a_n = [(9n+1)(9n)(9n-1)...(9n-(9n-1))] / (9n)!

a_n = (9n+1)

Now, we'll examine the limit as n approaches infinity:

lim (n -> ∞) a_n = lim (n -> ∞) (9n+1)

Since the term (9n+1) grows without bound as n approaches infinity, the limit does not exist. Therefore, the sequence diverges.

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The probability of drawing a green candy and a blue candy from the bag, one at a time and without replacement, is approximately 0.077 or 7.7%.

To calculate the probability of drawing a green candy and a blue candy from the bag without replacement, we need to determine the total number of candy combinations and the favorable combinations (green and blue candies).

The total number of candies in the bag is 3 blue + 4 green + 6 red = 13 candies.

When drawing the first candy, there are 13 candies to choose from. If a green candy is drawn, there are 4 green candies remaining out of the 12 candies left in the bag. Now, if a blue candy is drawn as the second candy, there are 3 blue candies remaining out of the 11 candies left in the bag.

The probability of drawing a green candy and a blue candy can be calculated as follows:

Probability = (Number of favorable combinations) / (Total number of candy combinations)

Number of favorable combinations = 4 (number of green candies) * 3 (number of blue candies) = 12

Total number of candy combinations = 13 (total candies) * 12 (remaining candies) = 156

Probability = 12 / 156 ≈ 0.077 or 7.7% (rounded to the nearest tenth)

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The limit lim(x→∞) x/√(x^2 + 3) is 1, and there is no need to apply L'Hospital's Rule in this case.

When trying to use L'Hospital's Rule to find the limit lim(x→∞) x/√(x^2 + 3), it is important to note that L'Hospital's Rule can only be applied if the function is continuous and differentiable. In this case, the function is continuous and differentiable, but applying L'Hospital's Rule is not necessary as the limit can be evaluated using another method.
First, let's rewrite the given function by dividing both the numerator and the denominator by x:
lim(x→∞) (x/x) / (√(x^2 + 3)/x) = lim(x→∞) 1 / √(1 + 3/x^2)
As x approaches infinity, the term 3/x^2 approaches 0, so the limit becomes:
lim(x→∞) 1 / √(1 + 0) = 1 / √(1) = 1
Therefore, the limit lim(x→∞) x/√(x^2 + 3) is 1, and there is no need to apply L'Hospital's Rule in this case.

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To calculate the height of the tree that would leave a 35 cm space from a 250 cm tall ceiling, subtract the desired space from the total height. Hence, the tree should be 215 centimeters tall.

The calculation involves understanding and using the concepts of height and ceiling space. When you want a tree to leave a certain amount of space from the ceiling, you have to subtract that desired space from the total height of the room. So, if your ceiling is 250 centimeters high and you desire a space of 35 centimeters to be left, the tree's height should be calculated as follows:

So, in this scenario, the tree must be 215 centimeters tall to fit perfectly under your ceiling with 35 centimeters of space left.

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[tex]\textit{area of a sector of a circle}\\\\ A=\cfrac{\theta r^2}{2} ~~ \begin{cases} r=radius\\ \theta =\stackrel{radians}{angle}\\[-0.5em] \hrulefill\\ r=6\\ \theta = \frac{2\pi }{3} \end{cases}\implies A=\cfrac{~~ \frac{2\pi }{3 }6^2 ~~}{2}\implies A=12\pi \stackrel{ using~\pi =3.14 }{\implies A=37.68}[/tex]