Test For each of the following scenarios, indicate which type of statistical error could have been committed or, alternatively, that no statistical CIP a. Unknown to the statistical analyst, the null hypothesis is actually true. OA. If the null hypothesis is rejected a Type I error would be committed. OB. If the null hypothesis is not rejected a Type I error would be committed. OC. If the null hypothesis is rejected a Type Il error would be committed. OD. If the null hypothesis is not rejected a Type Il error would be committed. OE. No error is made b. The statistical analyst fails to reject the null hypothesis OA. If the null hypothesis is true a Type I error would be committed. OB. If the null hypothesis is true a Type Il error would be committed OC. If the null hypothesis is not true a Type Il error would be committed OD. If the null hypothesis is not true a Type I error would be committed. OE. No error is made For each of the following scenarios, indicate which type of statistical error could have been committed or, alternatively, that no SCLOS c The statistical analyst rejects the null hypothesis. OA. If the null hypothesis is true a Type Il error would be committed OB. If the null hypothesis is not true a Type I error would be committed OC. If the null hypothesis is true a Type I error would be committed OD. If the null hypothesis is not true a Type Il error would be committed OE. No error is made d. Unknown to the statistical analyst, the null hypothesis is actually true and the analyst fails to reject the null hypothesis OA. A Type ll error has been committed. OB. Both a Type I error and a Type Il error have been committed OC. A Type I error has been committed OD. No error is made e Unknown to the statistical analyst, the null hypothesis is actually false I III = Test: Stat 11 For each of the following scenarios, indicate which type of statistical error could have been committed or, alternatively, that no statistical error w ACTE e. Unknown to the statistical analyst, the null hypothesis is actually false. OA. If the null hypothesis is not rejected a Type I error would be committed. OB. If the null hypothesis is rejected a Type I error would be committed. OC. If the null hypothesis is rejected a Type Il error would be committed OD. If the null hypothesis is not rejected a Type Il error would be committed OE. No error is made f Unknown to the statistical analyst, the null hypothesis is actually false and the analyst rejects the null hypothesis. OA. Both a Type I error and a Type Il error have been committed OB. A Type Il error has been committed. OC. A Type I error has been committed OD. No error is made

Answer:

14

Step-by-step explanation:

im right

Answer:

4.19 cubic feet

Step-by-step explanation:

The 90% confidence interval for the ratio of variances is [0.05, 12.36].

Lets calculate the sample variances for each polymer.

For Epoxy:

Sample Variance (s₁²) = (1.74² + 2.15² + 2.02² + 1.95²) / (4 - 1)

= 0.135

For MMA Prepolymer:

Sample Variance (s₂²) = (1.78² + 1.57² + 1.72² + 1.67²) / (4 - 1)

=0.056

F statistic is F = (s₁²) / (s₂²)

F=0.135/0.056

=2.41

Determine the degrees of freedom for each sample.

For Epoxy: df₁ = 4 - 1 = 3

For MMA Prepolymer: df₂ = 4 - 1 = 3

Now find the critical F-value corresponding to a 90% confidence level with df₁ and df₂ degrees of freedom.

Using statistical tables or a calculator, the critical F-value for a 90% confidence level with df₁ = 3 and df₂ = 3 is approximately 5.14.

Calculate the lower and upper bounds of the confidence interval.

Lower Bound = (1 / F) × (s₁² / s₂²)

= (1 / 5.14) × (s₁² / s₂²)

= 0.050

Upper Bound = 5.14 × (s₁² / s₂²)

= F × (s₁² / s₂²)

=12.36

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Answer:

12

Step-by-step explanation:

I know bc i did this before

Answer:

72 unit cubes (72 units³)

Step-by-step explanation:

[tex]6\times3\times4=72[/tex]

Answer:

B is your answer

Step-by-step explanation:

Answer:

the answer is the (L,N,K,O)

Answer:

slope: 3

y intercept:(0,-7)

Answer:

23/12 or 1 11/12

Step-by-step explanation:

Answer:

29/12 or 2 5/12

Step-by-step explanation:

Step-by-step explanation:

[tex]2 {v}^{2} + 14 = 104 \\ 2 {v}^{2} = 104 - 14 \\ 2 {v}^{2} = 90 \\ {v}^{2} = 90 \div 2 \\ {v}^{2} = 45 \\ v = \sqrt{45} [/tex]

I will leave the answer in square root form as i am not sure if you need to round your answer or not.

Given a prime number p and a polynomial f(x, y) of degree 2 with the only zero in Z/pZ being 0, we can show that f(ka, kb) = k^2f(a, b) using properties of polynomials.

i) To show that [tex]f(ka, kb) = k^2f(a, b)[/tex], we consider the polynomial f(x, y) and apply the properties of polynomials. Since f(x, y) has a degree of 2, we can write it as [tex]f(x, y) = ax^2 + bxy + cy^2,[/tex] where a, b, and c are coefficients. Now, substituting ka for x and kb for y, we get f(ka, kb) = [tex]a(ka)^2 + b(ka)(kb) + c(kb)^2[/tex]. Simplifying this expression, we obtain f(ka, kb) = [tex]k^2(ax^2 + bxy + cy^2) = k^2f(a, b),[/tex] which demonstrates the desired result.

ii) Using the result from part i), we can prove that if a is not congruent to 0 modulo p, then the equation f(x, y) ≡ a (mod p) always has a solution. Suppose f(x, y) ≡ a (mod p) has no solution for some value of a not congruent to 0 modulo p.  Therefore, if a is not congruent to 0 modulo p, we can choose an appropriate value of k such that [tex]k^2f(a, b)[/tex] ≡ b (mod p), leading to a solution for f(x, y) ≡ b (mod p). Thus, if a is not congruent to 0 modulo p, f(x, y) ≡ a (mod p) always has a solution.

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In a one-way ANOVA test with four groups, the null hypothesis states that there is no significant difference between the means of all four groups.

This means that any observed differences in the sample means are due to chance or random integral alone and not because of any systematic or real differences between the groups.

The null hypothesis assumes that the population means for each group are equal, which implies that there is no effect or influence of the independent variable on the dependent variable. If the null hypothesis is accepted, it means that any observed differences between the groups are not statistically significant and do not support the alternative hypothesis.

To determine whether to accept or reject the null hypothesis, researchers calculate the F-statistic, which compares the variability between the sample means to the variability within each group. If the calculated F-value is greater than the critical F-value for a given level of significance, the null hypothesis is rejected in favor of the alternative hypothesis, indicating that there is a significant difference between at least two of the group means.

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Answer:

The points which are reflections across both axis are;

1) (1.5, -2) and  [tex]\left (-1\dfrac{1}{2} , \ 2\right )[/tex]

2) (1.75, -4) and [tex]\left (-1\dfrac{3}{4} , \ 4\right )[/tex]

Step-by-step explanation:

The coordinate of the image of a point after a reflection across the 'x' and 'y' axis are given as follows;

[tex]\begin{array}{ccc}& Preimage&Image\\Reflection \ about \ the \ x-axis&(x, \ y)&(x, \, -y)\\Reflection \ about \ the \ y-axis&(x, \ y)&(-x, \, y)\end{array}[/tex]

Therefore, a reflection across both axis changes (only) the 'x' and 'y' value signs

The given points which are reflections across both axis are;

(1.5, -2) and  [tex]\left (-1\dfrac{1}{2} , \ 2\right )[/tex]

We note that [tex]\left (-1\dfrac{1}{2} , \ 2\right )[/tex] = (-1.5, 2)

The reflection of (1.5, -2) across the x-axis gives the image (1.5, 2)

The reflection of the image (1.5, 2) across the y-axis gives the image (-1.5, 2)

Similarly, we have;

(1.75, -4) and [tex]\left (-1\dfrac{3}{4} , \ 4\right )[/tex]

We note that [tex]\left (-1\dfrac{3}{4} , \ 4\right )[/tex] = (-1.75, 4)

The reflection of (1.75, -4) across the x-axis gives the image (1.75, 4)

The reflection of the image (1.75, 4) across the y-axis gives the image (-1.75, 4).

The other points have changes in the values of the 'x' and 'y' between the given pair and are therefore not reflections across both axis

Answer:

9.5 seconds.

Step-by-step explanation:

We know that the equation that represents the height of the penny as a function of time is:

h(t) = -16*t^2 + 1451

Where the height is in ft, and the time is in seconds.

a) We want to know how long takes the penny to hit the ground.

Well, the penny will hit the ground when its height is equal to zero.

Then we need to solve for t:

h(t) = 0 = -16*t^2 + 1451

0 = -16*t^2 + 1451

16*t^2 = 1451

t^2 = 1451/16 = 90.7

t = √90.7 = 9.5

This means that it takes 9.5 seconds to hit the ground.

Answer:9.5 seconds

Step-by-step explanation:you subtract

Answer:

15.9??

Step-by-step explanation:

Answer:

False. 2/3 is positive so therefore, it would be increasing, not decreasing.

Step-by-step explanation:

Answer:

c

Step-by-step explanation:

Answer:

C

Step-by-step explanation:

Answer:

the answer is minus 2 ( -2)

Answer:

B

Step-by-step explanation:

This involves reading the coordinates of points from the graph

When x = 3 then y = - 2 , that is the point (3, - 2 )

The Pythagorean Theorem is a concept in geometry that explains the relationship between the sides of a right-angled triangle. According to this theorem, the square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of its two legs. This can be represented in the form of an equation as c² = a² + b², where c is the hypotenuse and a and b are the legs of the triangle. This theorem is widely used in various fields, including architecture, engineering, and physics. It is named after the ancient Greek mathematician Pythagoras, who is credited with its discovery.

In simpler words, the Pythagorean theorem is used to find the length of the missing side of a right-angled triangle. For example, if we know the lengths of two sides of a triangle, we can use this theorem to find the length of the third side. This theorem is based on the fact that the hypotenuse is the longest side of a right-angled triangle, and that the square of a number is the product of that number multiplied by itself. It is also used to prove whether a triangle is a right-angled triangle or not. The Pythagorean Theorem is a fundamental concept in geometry and is an essential tool for solving various mathematical problems.

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Answer:

The class in the second and longest bar graph (the one labeled 3) has the most students.

Step-by-step explanation:

When looking for the largest amount of something in bar graphs, the largest bar graph is correct. In this case the second bar graph is the longest, and we can see it indicates the class contains 28 students.

Hypotheses and Claim:Null Hypothesis (H0): The median number of lottery tickets sold per day is 200.Alternative Hypothesis (HA): The median number of lottery tickets sold per day is below 200.

Claim: There is sufficient evidence to conclude that the median is below 200 tickets.

Critical Value(s):

To determine the critical value for the hypothesis test, we need to specify the significance level (α). In this case, α is given as 0.05.

Since the sample size is relatively small (n = 40), we can use the t-distribution to find the critical value. The critical value corresponds to the lower tail because we are testing whether the median is below 200 tickets.

Using a t-table or a statistical software, we find the critical value tα/2 with (n - 1) degrees of freedom. For α = 0.05 and (n - 1) = 39, we find t0.025 = -1.685.

Compute the Test Value:

To compute the test value, we need to calculate the test statistic, which is the t-value.

Let's define X as the number of days the owner sold fewer than 200 tickets. In this case, X follows a binomial distribution with n = 40 and p = 0.5 (assuming equal probability of selling more or fewer than 200 tickets).

Since the sample size is large enough, we can approximate the binomial distribution using the normal distribution. The mean (μ) and standard deviation (σ) of the binomial distribution can be calculated as follows:

μ = np = 40 * 0.5 = 20

σ = sqrt(np(1-p)) = sqrt(40 * 0.5 * 0.5) = sqrt(10)

The test statistic t is given by:

t = (X - μ) / (σ / sqrt(n))

In this case, X = 15, μ = 20, σ = sqrt(10), and n = 40. Plugging these values into the formula:

t = (15 - 20) / (sqrt(10) / sqrt(40)) ≈ -2.24

Make the Decision:

In this step, we compare the test value to the critical value.

If the test value falls in the rejection region (t < tα/2), we reject the null hypothesis. Otherwise, if the test value does not fall in the rejection region, we fail to reject the null hypothesis.

In our case, the test value t = -2.24 is smaller than the critical value tα/2 = -1.685.

Therefore, we reject the null hypothesis.

Summarize the Results:

Based on the analysis, there is sufficient evidence to conclude, at the α = 0.05 level, that the median number of lottery tickets sold per day is below 200.

The lottery outlet owner's hypothesis that she sells 200 lottery tickets a day is not supported by the data, indicating that the median sales are lower than the claimed value.

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Answer:

364×180=65,520 baskets

Step-by-step explanation:

since its a leap year(4days as shown in the document)

2nd question:

365×180=65,700 baskets

since its not a leap year

if u dont know leap years end with an even number which is 364 and non leap years end with odd number which is 365.

Answer:

Step-by-step explanation:

Answer:

9 square units

Step-by-step explanation:

Area of a square = base * height

= 3*3

= 9

Answer:

4 yards saved

Step-by-step explanation:

two adjacent sides: 6 + 8 = 14

diagonal: √(6² + 8²) = √100 = 10

14 - 10 = 4

Answer:

m = 0

Step-by-step explanation:

AB has a slope of y2 - y1 / x2 - x1 so -8-5/3-3 so -13/0 = undefined so it is a vertical line, so any line that is horizontal will be perpendicular to it, so horizontal lines have a zero slope

The interquartile range of the data set is 4.7 liters.

To find the interquartile range, we first need to sort the data from least to greatest. This gives us the following data set:

2.8, 7.5, 8.5, 11.6, 12, 12.1

The first quartile (Q1) is the median of the lower half of the data set. In this case, the lower half of the data set is {2.8, 7.5, 8.5}. The median of this data set is 7.5. Therefore, Q1 = 7.5.

The third quartile (Q3) is the median of the upper half of the data set. In this case, the upper half of the data set is {11.6, 12, 12.1}. The median of this data set is 12. Therefore, Q3 = 12.

The interquartile range (IQR) is calculated by subtracting Q1 from Q3. In this case, IQR = 12 - 7.5 = 4.7 liters.

The interquartile range is a measure of the variability of the middle 50% of the data. In this case, the interquartile range tells us that the middle 50% of the race car drivers have between 7.5 and 12 liters of gas in their tanks.

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a) The point on the curve where the tangent is horizontal is at t = 0.

b) The points where the tangent is vertical are at t₁ = -5 and t₂ = 5.

To find the points on the curve where the tangent is horizontal, we need to find the values of t that satisfy dy/dt = 0.

a.) Differentiating y = 3(t² - 75) with respect to t, we get:

dy/dt = 6t

Setting dy/dt = 0, we have:

6t = 0

t = 0

Therefore, when t = 0, the tangent is horizontal.

b.) To find the points where the tangent is vertical, we need to find the values of t that satisfy dx/dt = 0.

Differentiating x = t(t² - 75) with respect to t, we get:

dx/dt = 3t² - 75

Setting dx/dt = 0, we have:

3t² - 75 = 0

t² = 25

t = ±5

Therefore, the points where the tangent is vertical are when t = -5 and t = 5, with t₁ = -5 and t₂ = 5.

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The question is -

Consider the curve given by the parametric equations

x = t (t²-75) , y = 3 (t²-75)

a.) Determine the point on the curve where the tangent is horizontal.

t=

b.) Determine the points t_1,t_2 where the tangent is vertical and t_1 < t_2.

t_1=

t_2=

The confidence interval is (4.25, 6.75). If the df increases to 30 then the impact on the margin of error will decrease.

The test that is appropriate for the given values is the t-test since the sample size is small. The sample size is not given in the question but it is necessary to find the margin of error. Hence, the sample size can be found using the formula of the t-test. Let us recall the formula of the t-test below:t = (sample mean - population mean) / (sample standard deviation / √n)

Where,t = test statistic sample mean = 5.5population mean = 8sample standard deviation = 2.517confidence level = 95%corresponding test statistic value = 2.303Let's plug in all the values in the formula and solve for the sample size,n = ((sample mean - population mean) / (sample standard deviation/test statistic value))²= ((5.5 - 8) / (2.517 / 2.303))²= 3.484²≈ 12.11Hence, the sample size is 12.11 or approximately 12.

Now, let's calculate the margin of error using the formula below: Margin of error = (t-critical value) × (standard deviation / √n)Since the confidence level is 95%, the alpha level is 5% which is divided equally between the two tails, so the area in each tail is 2.5%. Using the t-distribution table with df = n-1 = 11 and alpha/2 = 0.025, the t-critical value is 2.201.

Let's plug in all the values in the formula of the margin of error and solve it: Margin of error = (t-critical value) × (standard deviation / √n)= 2.201 × (2.517 / √12)= 1.245 ≈ 1.25Therefore, the margin of error is 1.25.

The confidence interval can be calculated using the formula below: Confidence interval = (sample mean - margin of error, sample mean + margin of error)Confidence interval = (5.5 - 1.25, 5.5 + 1.25)= (4.25, 6.75)Hence, the confidence interval is (4.25, 6.75).If the df increases to 30 then the impact on the margin of error will decrease. If the pdf increases to 30, then the impact on the confidence interval will be narrow.

So, the confidence interval is (4.25, 6.75).If the df increases to 30 then the impact on the margin of error will decrease.

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