ASAP!!!! (Please do not copy off other people's answers.)

The following is a list of movie tickets sold each day for 10 days.

14, 35, 20, 23, 42, 87, 131, 125, 64, 92

Which of the following intervals are appropriate to use when creating a histogram of the data?

0 – 29, 30 – 59, 60 – 89, 90 – 119, 120 – 149
0 – 30, 30 – 55, 55 – 80, 80 – 105, 105 – 130
0 – 24, 25 – 49, 50 – 74, 75 – 99, 100 – 125
0 – 35, 35 – 70, 70 – 105, 105 – 140

To conduct a one-way analysis of variance (ANOVA) F-test at a significance level of α=0.05 to determine if the mean weight for Hispanic women of age 36 to 45 for four regions of the country are all equal, follow these steps:

Step:1. Download the data in your preferred format and import it into a statistical software (such as Excel, R, or SPSS).
Step:2. Perform the one-way ANOVA test using the software. The software will output the F-statistic, degrees of freedom for the numerator (df1), and degrees of freedom for the denominator (df2).
Step:3. Calculate the p-value using the software.
Step:4. Compare the p-value to the significance level (α=0.05) to make a decision and conclusion about the null hypothesis.
Without the actual data, I cannot provide specific results, but the process would look like this:
1. df1 = k - 1 (where k is the number of groups, in this case, 4 regions)
2. df2 = N - k (where N is the total number of observations)
3. Use the software to determine the F-statistic.
4. Calculate the p-value using the software.
Step:5. Compare the p-value to α=0.05:
  - If the p-value is less than 0.05, reject the null hypothesis and conclude that there is sufficient evidence that one or more of the mean weights for Hispanic women of age 36 to 45 are different.
  - If the p-value is greater than 0.05, fail to reject the null hypothesis and conclude that there is insufficient evidence to determine if the mean weights for Hispanic women of age 36 to 45 are different.

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Therefore, the proportionality equation and variable varies that can be used to find other combinations of b and c is: b = 50√c and Option (b) is correct: b = 50√c

We frequently use the phrase "a is proportional to b" when a directly fluctuates as b. When such is the case, a and b have the following algebraic relationship: a = kb. The proportionality constant is referred to as k. A relationship between a set of values for one variable and a set of values for other variables is known as a variation. direct change.

The function y = mx (commonly written y = kx), which is referred to as a direct variation, may be obtained from the equation y = mx + b if m is a nonzero constant and b = 0. Here b varies directly as the square root of c, we can write the equation as:

b = k√c

Here k is the constant of proportionality. To find the value of k, we can use the given values:

b = 100 when c = 4

100 = k√4

100 = 2k

k = 50

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The surface area of the figure is 132 square units.

Option D is the correct answer.

We have,

The figure has two types of shapes.

- 3 rectangles

- 2 triangles

Now,

Area of the 3 rectangles.

= 5 x 10 + 4 x 10 + 3 x 10

= 50 + 40 + 30

= 120 square units

Area of 2 triangles.

= 1/2 x 4 x 3 + 1/2 x 4 x 3

= 1/2 x 12 + 1/2 x 12

= 6 + 6

= 12 square units

Now,

Total surface area.

= 120 + 12

= 132 square units.

Thus,

The surface area of the figure is 132 square units.

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The Surface Area of the regular hexagonal prism is 92.784 square unit.

We have,

a = 2 unit

h= 6 unit

So, surface area of Prism

= 6 ah + 3√3 a²

= 6(2)(6) + 3√3 (2)²

= 72 + 12√3

= 92.784 square unit.

Thus, the Surface Area is 92.784 square unit.

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The coefficient of x³ term in the power series expansion of f(x) = [tex]e^x[/tex] sin(x) is 1/15.

To find the coefficient of x³ term in the power series expansion of f(x) = [tex]e^x[/tex] sin(x), we need to write the Taylor series for [tex]e^x[/tex] and sin(x) and then multiply them to get the Taylor series for f(x). The Taylor series for e^x is:

[tex]e^x[/tex] = 1 + x + (x²/2!) + (x³/3!) + ...

The Taylor series for sin(x) is:

sin(x) = x - (x³/3!) + (x⁵/5!) - ...

Multiplying these two series, we get:

f(x) = [tex]e^x[/tex] sin(x) = (1 + x + (x²/2!) + (x³/3!) + ...) × (x - (x³/3!) + (x⁵/5!) - ...)

Expanding this out and collecting the terms with x³, we get:

f(x) = x - (x³/3!) + (7x³/5!) + ...

Therefore, the coefficient of x³ term in the power series expansion of f(x) = [tex]e^x[/tex] sin(x) is -1/6 + 7/120 = 1/15.

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Answer:

Please mark me the brainliest

Bank Balance  | Amount Owed

---------------------|-------------

-$150               | $150

-$325              | $325

-$275              | $275

To find the amount owed, we simply take the absolute value of each bank balance. The player who owes the greatest amount is the one with the largest absolute value bank balance. In this case, that would be Malcolm, who owes $325.

Step-by-step explanation:

The trader's total sales for the month were GHC 280,000.0.

We can use the formula:

Commission = (Rate x Sales) / 100

where Commission is the amount of commission received, Rate is the commission rate, and Sales is the total sales made.

In this case, we are given:

Commission = GHC 35,000.0

Rate = 12.5%

Sales =?

Substituting these values into the formula, we get:

GHC 35,000.0 = (12.5 x Sales) / 100

Multiplying both sides by 100 and dividing by 12.5, we get:

Sales = (GHC 35,000.0 x 100) / 12.5

Simplifying, we get:

Sales = GHC 280,000.0

Therefore, the trader's total sales for the month were GHC 280,000.0.

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This was proved with advanced algebra that a doubled cube could never be constructed with a straightedge and compass. it is false.

It is a polygon having six faces. The volume of a cube is a side³

We have,

This statement is false.

Doubling the volume of a given cube will require increasing each side length by the cube root of 2.

However, this value is not constructible using only a straightedge and compass.

The Greeks were only able to construct lengths which could be expressed using a finite combination of rational numbers and square roots.

Thus,

This is not possible to construct a cube of twice the volume of a given cube using only a straightedge and compass.

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a) A basis for the solution space is the vector (3/4, 1, -1/4).

b) The dimension of the solution space is 1.

c) Basis for the row space is the vector (1, -2, 3, 2).

a) To find a basis for the solution space (nullspace) of the coefficient matrix, we can solve for the variables in terms of the free variable.

Starting with the augmented matrix [A|0]:

| 1  -2  3  2 |
| -3  6  -9  2 |

We can perform row operations to simplify the matrix:

R2 = R2 + 3R1

| 1  -2  3  2 |
| 0  0  0  8 |

Now, we can solve for the variables in terms of the free variable:

x - 2y + 3z = -2z

z = -1/4t
y = t
x = 3/4t

So the solution space can be written as:

t * (3/4, 1, -1/4)

Thus, a basis for the solution space is the vector (3/4, 1, -1/4).

b) The dimension of the solution space (nullity) is the number of free variables, which in this case is 1.

So the dimension of the solution space is 1.

c) To find a basis for the row space of the coefficient matrix, we can row reduce the matrix and take the non-zero rows as a basis.

Starting with the augmented matrix [A|0]:

| 1  -2  3  2 |
| -3  6  -9  2 |

We can perform row operations to simplify the matrix:

R2 = R2 + 3R1

| 1  -2  3  2 |
| 0  0  0  8 |

The row space is spanned by the non-zero rows of the row reduced matrix:

(1, -2, 3, 2)

So a basis for the row space is the vector (1, -2, 3, 2).

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Check the picture below.

so if we just get the volume of the whole box, and the volume of the balls, if we subtract the volume of the balls from that of the whole box, what's leftover is the part we didn't subtract, namely the empty space.

[tex]\stackrel{ \textit{\LARGE volumes} }{\stackrel{ whole~box }{(3.5)(3.5)(12.1)}~~ - ~~\stackrel{\textit{three balls} }{3\cdot \cfrac{4\pi (1.65)^3}{3}}} \\\\\\ 148.225~~ - ~~17.9685\pi ~~ \approx ~~ \text{\LARGE 91.8}~cm^3[/tex]

To set up this confidence interval, first identify the population parameter of interest, next select the appropriate estimator, then check the conditions for constructing the confidence interval that are: Randomization, Sample size and Distribution shape.

State:
Zachary wants to estimate the mean number of daily texts he sent over the past four years using a 90% confidence interval. He has an SRS of 50 days, with a daily average of 22.5 messages. The data is strongly skewed to the right with many outliers.

Plan:
1. Identify the population parameter of interest: The mean number of daily texts sent by Zachary over the past four years (µ).
2. Select the appropriate estimator: In this case, it's the sample mean = 22.5 messages.
3. Check the conditions for constructing the confidence interval:
  a. Randomization: Zachary used a simple random sample (SRS) of 50 days, which satisfies the randomization condition.
  b. Sample size: The sample size is n = 50, which is typically considered large enough for constructing a confidence interval.
  c. Distribution shape: Since the data is strongly skewed to the right with many outliers, the normality condition might not be satisfied. In this case, the Central Limit Theorem (CLT) may not apply, and the confidence interval might not be accurate.

Given the potential issue with the distribution shape, Zachary should consider either transforming the data to approximate normality or using a nonparametric method.

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The average rate of change of f(x) = x with respect to x over the interval a = 1 is 1.

To determine the average rate of change of f(x) = x with respect to x over the interval a, we'll use the formula:

Average Rate of Change = (f(b) - f(a)) / (b - a)

In this case, the interval a is 1, so let's choose an interval b. We can use any value for b, but let's choose b = 2 for simplicity.

Step 1: Find f(a) and f(b)
f(x) = x, so:
f(1) = 1
f(2) = 2

Step 2: Plug the values into the formula
Average Rate of Change = (f(2) - f(1)) / (2 - 1)
Average Rate of Change = (2 - 1) / (2 - 1)

Step 3: Calculate the result
Average Rate of Change = (1) / (1)

The average rate of change of f(x) = x with respect to x over the interval a = 1 is 1.

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There is no solution to this problem. None of the answer choices (A, B, C, D) are correct.

Let's start by representing the original two-digit number as 10x + y, where x represents the tens digit and y represents the ones digit.

When we reverse the digits, we get the number 10y + x. According to the problem, this number is 9 less than three times the original number:

10y + x = 3(10x + y) - 9

Simplifying this equation, we get:

10y + x = 30x + 3y - 9

7y - 29x = -9

We also know that the difference between these two numbers is 45:

(10x + y) - (10y + x) = 45

9x - 9y = 45

x - y = 5

Now we have two equations with two variables, which we can solve using substitution or elimination. I'll use elimination:

7y - 29x = -9
-7y + 7x = 35 (multiplying the second equation by -7)

Adding these two equations, we get:

-22x = 26

x = -13/11

This doesn't make sense, since x should be a digit between 1 and 9.

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Option e. "The number of trials is not fixed" would be the correct answer.

The assumption of the Binomial distribution that is NOT included in the options provided is that the number of trials must be fixed in advance. This means that the Binomial distribution applies only to situations where there is a fixed number of independent trials, each with the same probability of success, and the interest is in the number of successes that occur in these trials. Therefore, option e. "The number of trials is not fixed" would be the correct answer.

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Answer:

dark blue

Step-by-step explanation:

We do not have sufficient evidence to conclude that the variance of monthly incomes is higher for male employees than it is for female employees at the 0.01 level of significance.

a) State the hypotheses:

Null Hypothesis (H0): The variance of monthly incomes for male employees is equal to or less than the variance of monthly incomes for female employees.

Alternative Hypothesis (Ha): The variance of monthly incomes for male employees is higher than the variance of monthly incomes for female employees.

b) Calculate the test statistic:

We can use the F-test to compare the variances of the two samples. The test statistic is:

[tex]F = s1^2 / s2^2[/tex]

where s1 and s2 are the sample standard deviations, and F follows an F-distribution with (n1-1) and (n2-1) degrees of freedom.

For female employees:

n1 = 61

[tex]s1 = $194.40[/tex]

[tex]s1^2 = ($194.40)^2 = $37,825.60[/tex]

For male employees:

n2 = 121

s2 = $269.92

[tex]s2^2 = ($269.92)^2 = $72,941.29[/tex]

So, the test statistic is:

[tex]F = s1^2 / s2^2 = $37,825.60 / $72,941.29 = 0.518[/tex]

c) State the rejection criterion for the null hypothesis:

We will use a significance level of 0.01. Since this is a one-tailed test (we are testing if the variance of male employees is higher than the variance of female employees), the rejection region is in the upper tail of the F-distribution. We need to find the critical value of F with (60, 120) degrees of freedom at the 0.01 level of significance. Using a statistical table or calculator, we find that the critical value is 2.74.

d) Draw your conclusion:

The calculated F-value (0.518) is less than the critical F-value (2.74). Therefore, we fail to reject the null hypothesis. We do not have sufficient evidence to conclude that the variance of monthly incomes is higher for male employees than it is for female employees at the 0.01 level of significance.

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The density of the ball in the air is given as follows:

d = 4 x 10^5 ounces/ft³.

The density is calculated as the division of the mass by the volume of an object, as follows:

d = m/v.

The ball in this problem is spherical with a diameter of 0.05 feet = radius of 0.025 feet, hence the volume is given as follows:

V = 4 x 3.1416 x 0.025³/3

V = 6.545 x 10^-5 ft³.

The ball in the air is inflated, hence the mass is given as follows:

m = 22.93 ounces.

Thus the density of the ball is given as follows:

d = 22.93/(6.545 x 10^-5)

d = 4 x 10^5 ounces/ft³.

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After the battery is fully charged, Car B can go further than Car A.  Car B, as compared to Car A, had lower variability measurements. After the battery is completely charged, Car B can go further than Car A since Car A has a lower mean and median. Option D is Correct.

The median splits the data in half. A lower median indicates that Car A has less mileage than Car B.

Two measurements exist.

The measure of centre reveals how closely or widely the data are dispersed around the centre.

The measurements of centre are mean, median, and mode.

Car A travelled less since it had a lower mean and median.

We can find out how data changes with a single value using the measure of variability. The data is denser at the mean when the MAD is less. The MAD in Car B is lower. Data that is closer to the centre of the data set has a smaller IQR.

IQR is lower in Car B.

Consequently, automobile B travelled steadily since its IQR and MAD were lower. Option D is Correct.

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Correct Question:

Annabel is comparing the distances that two electric cars can travel after the battery is fully charged. Car A (miles) Car B (miles) Mean 145 200 Median 142 196 IQR 8 4 MAD 6 2 Part A Use the measures of center to make an inference about the data. Use the drop-down menus to complete your answer. Car A can travel further than Car B after the battery is fully charged. Part B Based on the data, which car performs most consistently? Explain. A. Car A because the measures of center are smaller for Car A than for Car B. B. Car B because the measures of center are smaller for Car B than for Car A. C. Car A because the measures of variability are smaller for Car A than for Car B. D. Car B because the measures of variability are smaller for Car B than for Car A.

Answer:

Step-by-step explanation:

To determine if there is a positive linear association between the hours worked and the daily wages of waiters in a restaurant, you can create a scatter plot of the data and look for a pattern.

Once you have the data, you can use a statistical software or a spreadsheet program to create a scatter plot. You can then visually inspect the scatter plot to see if there is a clear pattern of a positive linear association between the two variables.

If there is a positive linear association, the data points on the scatter plot will form a roughly straight line that slopes upwards from left to right. The closer the data points are to the line, the stronger the association.

So, the data table that indicates a positive linear association between the hours worked and the daily wages of waiters in a restaurant is the one where the scatter plot shows a clear upward trend.

For the first differential equation, dx/dy = 5x/y, we can separate the variables and integrate:

dy/dx = y/5x
(1/y)dy = (1/5x)dx
Integrating both sides, we get:
ln|y| = (1/5)ln|x| + C
where C is the constant of integration.

To solve for y, we can exponentiate both sides:
|y| = e^(ln|x|/5 + C)
|y| = Ce^(ln|x|/5)
where C is a constant of integration.

Since we don't know whether x and y are positive or negative, we can write the general solution as:
y = ± Cx^(1/5)

For the second differential equation, dx/dy = x^4, we can again separate the variables and integrate:

dy/dx = 1/x^4
x^4dy = dx
Integrating both sides, we get:
(1/3)x^3y = x + C
where C is the constant of integration.

To solve for y, we can multiply both sides by (3/x^3):
y = (3/x^3)(x + C)
y = 3/x^2 + 3Cx^(-3)

So the general solution to the differential equation dx/dy = x^4 is:
y = 3/x^2 + 3Cx^(-3), where C is a constant of integration.

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Answer:

OD

Step-by-step explanation:

Angles in a triangle add to 180 degrees. The only set of angles which total to 180 is the values in OD

The second Borel-Cantelli lemma, we have P(lim sup Sn < ∞) = 0, which implies that Sn converges almost surely.

Let {Xn} be a sequence of random variables, and let Sn = X1 + X2 + ... + Xn be the corresponding sequence of partial sums. We want to show that if E(Xn²) is finite for all n, then Sn converges almost surely.

Let Yn = Xn^2. Then E(Yn) = E(Xn²) < ∞ for all n, since we are given that the second moments are finite. By the second Borel-Cantelli lemma, it suffices to show that the series ∑ P(Yn > ε) converges for every ε > 0.

Since Yn = Xn² ≥ 0, we have P(Yn > ε) ≤ P(|Xn| > √ε). Using Markov's inequality, we have:

P(|Xn| > √ε) ≤ E(|Xn|²)/ε = E(Yn)/ε.

Therefore, we have:

∑ P(Yn > ε) ≤ ∑ E(Yn)/ε = (1/ε) ∑ E(Yn) = (1/ε) ∑ E(Xn²) < ∞.

The last inequality follows from the fact that the second moments are assumed to be finite.

Thus, by the second Borel-Cantelli lemma, we have P(lim sup Sn < ∞) = 0, which implies that Sn converges almost surely.

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Result:

1. Based on the properties, the most precise name for figure is A. parallelogram

2. From the properties, the most precise name for the figure is B. rhombus.

We can determine the precise name of the figure calculating the slopes of AB, BC, CD, and DA using the slope formula and/or the distance formula:

1. Using the slope formula:

AB = (0 - (-6))/(2 - (-5)) = 2

BC = (9 - 0)/(11 - 2) = 9/9 = 1

CD = (3 - 9)/(4 - 11) = -6/-7 = 6/7

DA = (-6 - (-5))/( -5 -(-5)) = 0

Calculate the lengths of the sides using distance formula:

AB = [tex]\sqrt((2 - (-5))^2 + (0 - (-6))^2)[/tex] = [tex]\sqrt(7^2 + 6^2)[/tex] = [tex]\sqrt{85}[/tex]

f BC = [tex]\sqrt((11 - 2)^2 + (9 - 0)^2)[/tex] = [tex]\sqrt(9^2 + 9^2)[/tex] = 9√2)

CD = [tex]\sqrt((4 - 11)^2 + (3 - 9)^2)[/tex] = sqrt[tex]\sqrt(7^2 + 6^2)[/tex] = √85

DA = [tex]\sqrt((-5 - 4)^2 + (-6 - (-9))^2)[/tex] = [tex]\sqrt(9^2 + 3^2)[/tex] = 3√10

The slopes of AB and CD are equal (2 and 6/7, respectively), and the slopes of BC and DA are equal (1 and 0, respectively).

Therefore, opposite sides are parallel that is a parallelogram.

2. First, we can calculate the slopes of AB, BC, CD, and DA using the slope formula:

AB = (1 - (-4))/(-7 - (-9)) = 5/2

BC = (5 - 1)/(1 - (-7)) = 4/4 = 1

CD = (0 - 5)/(-1 - 1) = -5/-2 = 5/2

DA = (-4 - 0)/(-9 - (-1)) = 4/8 = 1/2

Next, using the distance formula, we calculate the lengths of the sides:

AB = [tex]\sqrt((-7 - (-9))^2 + (1 - (-4))^2)[/tex] = [tex]\sqrt(2^2 + 5^2)[/tex] = [tex]\sqrt29[/tex]

BC = [tex]\sqrt{(1 - (-7))^2 + (5 - 1)^2}[/tex] = [tex]\sqrt(8^2 + 4^2)[/tex] = 4[tex]\sqrt17[/tex]

CD = [tex]\sqrt((-1 - 1)^2 + (0 - 5)^2)[/tex] = [tex]\sqrt(2^2 + 5^2)[/tex] = [tex]\sqrt29[/tex]

DA = [tex]\sqrt((-9 - (-1))^2 + (-4 - 0)^2)[/tex] = [tex]\sqrt(8^2 + 4^2)[/tex] = [tex]\sqrt80[/tex])

The slopes of AB and CD are equal (5/2 and 5/2, respectively), and the slopes of BC and DA are equal (1 and 1/2, respectively). meaning the opposite sides are parallel.

AB and CD have the same length ([tex]\sqrt(29)[/tex]), and BC and DA have the same (4[tex]\sqrt(17}[/tex]), which means it's a rhombus.

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The equation which can be used to determine the total amount, t, Frankie will have spent after m months on his cell phone plan is t = 30 + 45m.

Given that,

Frankie has a new cell phone plan.

He will pay a one-time activation fee of 30$, and 45$ each month.

One time activation fee = $30

Amount each month = $45

Amount for m months = 45m

Total amount for the plan = 30 + 45m

If t represents the total amount for the cell phone activation plan, the required equation can be written as,

t = 30 + 45m

Hence the required equation for the cell phone plan is t = 30 + 45m.

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When a=0.02 and n=24, [tex]X_{left}^{2}[/tex] = 9.260 and [tex]X_{right}^{2}[/tex]= 41.638. In order to calculate [tex]X_{left}^{2}[/tex] and [tex]X_{right}^{2}[/tex] when a=0.02 and n=24, we need to use the chi-squared distribution table. This table provides us with the critical values for a given level of significance (alpha) and degrees of freedom (df).

To answer your question, when a=0.02 and n=24, we will find the [tex]X_{left}^{2}[/tex] and [tex]X_{right}^{2}[/tex] values using the Chi-square distribution table.

Step 1: Determine the degrees of freedom. In this case, the degrees of freedom (df) are equal to n-1, so df = 24 - 1 = 23.

Step 2: Determine the significance level (alpha) and divide it by 2. Since a = 0.02, the significance level is [tex]\frac{\alpha}{2} =0.01[/tex] for each tail (left and right) of the distribution.

Step 3: Use the Chi-square distribution table to find the critical values. Look for the values corresponding to the degrees of freedom (23) and significance level (0.01) in each tail.

According to the Chi-square distribution table:
[tex]X_{left}^{2}[/tex]= 9.260
[tex]X_{right}^{2}[/tex]= 41.638

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This change in shipping costs may or may not affect the company.

We need additional information to determine the exact effect.

Option B is the correct answer.

We have,

While the reduction in shipping costs from Plant 3 to the North region is significant, we need more information about the company's current shipping plan, routes, and costs associated with other plants to determine if this change will impact their overall shipping strategy.

Thus,

This change in shipping costs may or may not affect the company.

We need additional information to determine the exact effect.

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The value of θ from the given right triangle is 50 degree.

The legs of given right angle triangle are 6 units and 5 units.

Here, opposite side = 6 units and adjacent side = 5 units

We know that, tanθ= Opposite/Adjacent

tanθ= 6/5

tanθ= 1.2

θ=50.19

θ≈50°

Therefore, the value of θ from the given right triangle is 50 degree.

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The answer is not one of the choices given. The closest choice is (e) [74.6 - 88.1] and 94.8, but the midpoint of the fourth class is actually 95.15, not 94.8.

To find the class interval, we first need to calculate the range of the data:

Range = maximum observation - minimum observation

Range = 128.4 - 47.6

Range = 80.8

Next, we need to determine the width of each class interval:

Width of each class interval = Range / Number of classes

Width of each class interval = 80.8 / 6

Width of each class interval ≈ 13.47 ≈ 13.5 (rounded to one decimal place)

Now we can determine the class intervals:

1st class: 47.6 - 61.1

2nd class: 61.2 - 74.7

3rd class: 74.8 - 88.3

4th class: 88.4 - 101.9

5th class: 102.0 - 115.5

6th class: 115.6 - 129.1

So the third class is [74.8 - 88.3] and the midpoint of the fourth class is:

Midpoint of the fourth class = Lower limit of the fourth class + (Width of each class interval / 2)

Midpoint of the fourth class = 88.4 + (13.5 / 2)

Midpoint of the fourth class = 88.4 + 6.75

Midpoint of the fourth class = 95.15

Therefore, the answer is not one of the choices given. The closest choice is (e) [74.6 - 88.1] and 94.8, but the midpoint of the fourth class is actually 95.15, not 94.8.

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Answer:

1/870 ≈ 0.0011494 or about 0.115% (rounded to 6 decimal places)

Step-by-step explanation:

The first digit can be any of the numbers between 10 and 40, except for those that end in 6 (since the last digit of all three numbers is 6). This leaves us with 30 numbers to choose from for the first digit. Similarly, the second digit can be any of the numbers between 10 and 40, except for those that end in 6 and the one chosen for the first digit. This leaves us with 29 numbers to choose from for the second digit.

For the third digit, we have only one option since we know it ends in 6.

So the total number of possible combinations is:

30 * 29 * 1 = 870

Out of these, only one combination is the correct one. Therefore, the probability of guessing the combination correctly on the first try is:

1/870 ≈ 0.0011494 or about 0.115% (rounded to 6 decimal places)