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daser333 [38]
1 year ago
15

Eva earned the following test scores in science: 82, 100, 96, 90, 91 Ms. Wade entered the next score as 0 because Eva was absent

on test day and hadn’t had a chance to make up the test. Which statements are true based on the data? Check all that apply. After adding the 0 test score, the mean would be the most appropriate measure of center to describe the data. After adding the 0 test score, the mean would be affected. After adding the 0 test score, the median would be the most appropriate measure of center to describe the data. Before the missed test, Eva’s median score was 96. Before the missed test, Eva’s median score was 91. Before the missed test, Eva’s mean score was 91.8.​

Mathematics
2 answers:
xz_007 [3.2K]1 year ago
7 0

Answer:

2,3,5,6

Step-by-step explanation:

slega [8]1 year ago
3 0

Answer:

a) After adding the 0 test score, the mean would be the most appropriate measure of center to describe the data-------> is true

b) After adding the 0 test score, the mean would be affected-----> is true

c) After adding the 0 test score, the median would be the most appropriate measure of center to describe the data-----> is false

d) Before the missed test, Eva’s median score was 96-----> is false

e) Before the missed test, Eva’s median score was 91------ is true

f) Before the missed test, Eva’s mean score was 91.8-----> is true

Step-by-step explanation:

Hope this helps:)

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Marina86 [1]

Answer:

A

C

D

E

Step-by-step explanation:

Exterior angles can be described as the angles that are formed between the side of a polygon and the extended adjacent side of the polygon.

Or an exterior angle is the angle that is not inside the triangle formed.

The angles inside the triangle are interior angles.

Exterior angles are :

2

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6 0
2 years ago
In a certain classroom ,12.5% of the students own at least one pet. If there are 32 students in the classroom,how many students
Y_Kistochka [10]

Answer:

28

Step-by-step explanation:

Given that 12.5% of students own at least 1 pet, then the percentage of students who do not own a pet is

100% - 12.5% = 87.5%

Calculate 87.5% of 32, that is

\frac{87.5}{100} × 32

= 0.875 × 32 = 28 ← number of students who do not own a pet

8 0
2 years ago
Read 2 more answers
Consider the discussion in our Devore reading in this unit involving an important distinction between mean and median that uses
Levart [38]

Answer:

Step-by-step explanation:

A trimmed mean is a method of averaging that removes a small designated percentage of the largest and smallest values before calculating the mean. After removing the specified observations, the trimmed mean is found using a standard arithmetic averaging formula. The use of a trimmed mean helps eliminate the influence of data points on the tails that may unfairly affect the traditional mean.

trimmed means provide a better estimation of the location of the bulk of the observations than the mean when sampling from asymmetric distributions;

the standard error of the trimmed mean is less affected by outliers and asymmetry than the mean, so that tests using trimmed means can have more power than tests using the mean.

if we use a trimmed mean in an inferential test , we make inferences about the population trimmed mean, not the population mean. The same is true for the median or any other measure of central tendency.

I can imagine saying the skewness is such-and-such, but that's mostly a side-effect of a few outliers, the fact that the 5% trimmed skewness is such-and-such.

I don't think that trimmed skewness or kurtosis is very much used in practice, partly because

If the skewness and kurtosis are highly dependent on outliers, they are not necessarily useful measures, and trimming arbitrarily solves that problem by ignoring it.

Problems with inconvenient distribution shapes are often best solved by working on a transformed scale.

There can be better ways of measuring or more generally assessing skewness and kurtosis, such as the method above or L-moments. As a skewness measure (mean ? median) / SD is easy to think about yet often neglected; it can be very useful, not least because it is bounded within [?1,1][?1,1].

i expect to see the optimum point in that process at some value between the mean and median.

3 0
2 years ago
On a test, leo is asked to completely factor the polynomial 3x3 – 3x 5x2 – 5. he uses double grouping to get (x2 – 1)(3x 5). has
Agata [3.3K]

No, Leo's answer is not a product of prime polynomials because x2 – 1 can be factored. This is a difference of squares. He should continue factoring to get

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8 0
2 years ago
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Which of the following statements are true concerning the mean of the differences between two dependent samples​ (matched pairs)
elena-s [515]

Answer:

A and C

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations we can use it.  

And in order to conduct it we need to have some assumptions:

1) The dependent variable must be continuous (interval/ratio).

2) The observations are independent of one another.

3) The dependent variable should be approximately normally distributed.

4) The dependent variable should not contain any outliers.

Let's analyze one by one the options on this case:

A. The methods used to evaluate the mean of the differences between two dependent variables apply if one has 86 IQ scores of taxpayers from Texas and 86 IQ scores of taxpayers from Ohio

We can assume that is true since the two variables are dependent and we are assuming that all the other conditions are satisfied to use the t paired test.

B. The requirement of a simple random sample is satisfied if we have matched pairs of voluntary response data.

That's not neccesary true since is not a requirement in order to use the t pairedtest.

C. If one has more than 23 matched pairs of sample​ data, one can consider the sample to be large and there is no need to check for normality.

We can assume that is true since we need to ensure normality in order to apply the test and if the sample size is large enough large we can apply the test.

D. If one has twenty matched pairs of sample​ data, there is a loose requirement that the twenty differences appear to be from a normally distributed population.

False the requirement of normality is important to apply the test not a loose requirement.

E. If one wants to use a confidence interval to test the claim that mu Subscript d Baseline greater than 0 with a 0.01 significance​ level, the confidence interval should have a confidence level of 98​%.

That's not correct the significance level is \alpha=1-0.98=0.02 and that not correspond to the significance level given on this case.

7 0
2 years ago
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