Read the following excerpt from The Age of Innocence, and identify the narrator’s tone regarding Countess Olenska’s behavior.



"To the general relief, the Countess Olenska was not present in her grandmother's drawing-room during the visit of the betrothed couple. Mrs. Mingott said she had gone out; which, on a day of such glaring sunlight, and at the "shopping hour," seemed in itself an indelicate thing for a compromised woman to do."

confusion
relief
annoyance
disapproval

Answers

Answer 1

Answer:

the tone is confusion

you can tell because the narrator is quite perplexed by why the Countess was not present in her grandmother's room, and why she had gone out during the sunlight at noon ("shopping hour") , since it is strange for her a married woman like her to do so.

Answer 2
The answer would be the first one

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How could you describe the value of using a correlation to examine your data?

Answers

Answer:

Correlation is a statistical technique that can show whether and how strongly pairs of variables are related. For example, height and weight are related; taller people tend to be heavier then start short people. Correlation can tell you just how much of the variations in peoples weights is related to their heights.

The way I will describe the value of a data using a correlation is that; The correlation data gives us the relationship between the factors considered.

What are the uses of Correlation?

Correlation is a term in statistics that is used to test the relationships between quantitative variables or categorical variables. This means that Correlation, is a measure of how things are related.

The study of how variables are correlated is called correlation analysis.

Correlations are very useful because if you can find out what relationship variables have, you can make predictions about future behavior.

Read more about Statistics Correlation at; https://brainly.com/question/4219149

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Answer:

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Explanation:

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Suppose the National Transportation Safety Board (NTSB) wants to examine the safety of compact cars, midsize cars, and full-size cars. It collects a sample of three for each of the treatments (cars types). Using the hypothetical data provided below, test whether the mean pressure applied to the driver’s head during a crash test is equal for each types of car. Use α = 5%. Table ANOVA.1 Compact cars Midsize cars Full-size cars 643 469 484 655 427 456 702 525 402 X 666.67 473.67 447.33 S 31.18 49.17 41.68

Answers

Answer:

Explanation:

5320 5402 4064 5652

The mean pressure applied to the driver's head during a crash test is not equal for at least one pair of car types (compact, midsize, and full-size cars).

How did we arrive at this assertion?

To test whether the mean pressure applied to the driver's head during a crash test is equal for each type of car (compact, midsize, and full-size), we can use a one-way analysis of variance (ANOVA). ANOVA is used to compare the means of three or more groups.

First, let's define the null and alternative hypotheses:

Null hypothesis (H₀): The mean pressure applied to the driver's head is equal for compact, midsize, and full-size cars.

Alternative hypothesis (H₁): The mean pressure applied to the driver's head is not equal for at least one pair of car types.

We will use a significance level (α) of 5% (0.05).

Given the data:

Compact cars: 643, 655, 702

Midsize cars: 469, 427, 525

Full-size cars: 484, 456, 402

Let's calculate the necessary statistics for the ANOVA table:

Compact cars:

Sample mean (X): 666.67

Sample standard deviation (S): 31.18

Midsize cars:

Sample mean (X): 473.67

Sample standard deviation (S): 49.17

Full-size cars:

Sample mean (X): 447.33

Sample standard deviation (S): 41.68

Now, we can proceed with the ANOVA test and calculate the F-statistic and p-value.

The ANOVA table looks as follows:

Source of Variation | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F-statistic

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

Between Groups | SS_between | k - 1 | MS_between = SS_between / (k - 1) | F = MS_between / MS_within

Within Groups | SS_within | N - k | MS_within = SS_within / (N - k) |

Total | SS_total | N - 1 |

Where:

- k is the number of groups (car types)

- N is the total number of observations

- SS represents the sum of squares

Let's calculate the ANOVA table using the provided data:

Total number of observations (N): 3 groups x 3 observations per group = 9

Number of groups (k): 3

Between Groups:

Mean of all observations (grand mean):

x-cap = (666.67 + 473.67 + 447.33) / 3 = 529.89

SS_between = (3 × ((666.67 - 529.89)² + (473.67 - 529.89)² + (447.33 - 529.89)²))

Within Groups:

SS_within = (2 × (31.18² + 49.17² + 41.68²))

Total:

SS_total = SS_between + SS_within

Now, we can calculate the degrees of freedom (df) for each source of variation:

df_between = k - 1 = 3 - 1 = 2

df_within = N - k = 9 - 3 = 6

df_total = N - 1 = 9 - 1 = 8

Finally, we can calculate the mean squares:

MS_between = SS_between / df_between = 343,640.67 / 2 = 171,820.335

MS_within = SS_within / df_within = 32,533.99 / 6 = 5,422.33167

We have obtained the actual values for the ANOVA calculations. Now we can proceed to calculate the F-statistic and determine whether to reject or fail to reject the null hypothesis.

Now, let's calculate the F-statistic:

F = MS_between / MS_within = 171,820.335 / 5,422.33167 ≈ 31.73

To determine whether to reject or fail to reject the null hypothesis, we need to compare the calculated F-statistic to the critical F-value from the F-distribution table at the given significance level (α = 0.05). However, we need to determine the critical F-value based on the degrees of freedom for the numerator (df_between) and denominator (df_within).

For df_between = 2 and df_within = 6, the critical F-value at α = 0.05 is approximately 5.14.

Since the calculated F-statistic (31.73) is greater than the critical F-value (5.14), we can reject the null hypothesis.

Therefore, we conclude that the mean pressure applied to the driver's head during a crash test is not equal for at least one pair of car types (compact, midsize, and full-size cars).

learn more about ANOVA: https://brainly.com/question/25800044

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At what stage of the Model of Situational Leadership does shared decision making occur while simultaneously less direction is provided?
Group of answer choices

adjourning

telling

selling

participating

delegating

Answers

Answer:

The answer is participlating.

I apologize if I am incorrect.

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5) In the above experiment, describe how to conduct it as a double-blind study.

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