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rjkz [21]
1 year ago
14

Check the following data set for outliers if there is more than one answer separate them with the commas

Mathematics
1 answer:
ch4aika [34]1 year ago
8 0

We have the following data set: {245, 253, 220, 241, 282, 234, 236, 249}

The mean value is 245.

The standard deviation is 18 (rounded)

An outlier is any data that diverges significantly from the majority of the data set.

If we consider as outliers every data that distances from the mean in more than one standard deviation, then the outliers will be any data less than 227 or greater than 263

Therefore, the outliers are 220 and 282.

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Can 12/100 be simplified
Alborosie
Yes it can be simplified and the answer is 3/25

how? 
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7 0
3 years ago
Let z be a bernoulli random variable with parameter p and w an independent binomial random variable with parameters n and p. Wha
Maurinko [17]

Answer:

It is known that if Z is a binomial random variable with parameters n  and p and in addition, W is a binomial random variable, independent of X, with parameters m and p, it is assumed that the variable R = Z + W, is a binomial random variable with parameters (n + m and p. In this case, m = 1. Therefore, R is a binomial random variable with parameters (n + 1) and p.

Step-by-step explanation:

4 0
3 years ago
Explain why figure b is not the image of figure a after a reflection using line l
Kaylis [27]

Answer:

See below

Step-by-step explanation:

Figure A has a square cut out into it as figure B has a trapezoid cut out into it.

A reflection is mirroring an image, this means that the image must be identical to its reflected image.

Because figure A and B have different shapes cut into it they are not identical and can therefore not be reflections.

8 0
3 years ago
Read 2 more answers
Look at pic!! #9 pls. Thank youuuuu. 20 points
Karolina [17]
35n=280
n=8 8 dimes and 8 quarters
4 0
3 years ago
Information on a packet of​ seeds, which may not be a random sample of​ seeds, claims that the germination rate is 94​%. ​What's
Evgesh-ka [11]

Answer:

0.1052

Step-by-step explanation:

Given that proportion of germination in the population is 94% =0.94

p = 0.96

Sample size = 220

Std dev of p = \sqrt{\frac{pq}{n} } \\=0.016

The probability that more than 96​% of the 220 seeds in the packet will​ germinate

= P(p\geq 0.96)\\=P(Z\geq \frac{0.96-0.94}{0.016})\\=P(Z\geq 1.25)\\==1-0.8948\\=0.1052

Assumptions are np and nq >5 and also sample size >220 hence normal

7 0
3 years ago
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