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3 years ago

The mean of a set is not affected by outliers. true or false.

Mathematics

2 answers:

Yuri [45]3 years ago

8 0

It is false beacuse you have to add up all the numbers including the outliers to get the mean

elena55 [62]3 years ago

6 0

Answer:

False.

Step-by-step explanation:

We have been given a statement. We are asked to determine whether our given statement is true or false.

Statement:

The mean of a set is not affected by outliers.

We know that mean is the average of all the data points of a data set.

When our data set will have a large value data point, the mean will be above the center of data set.

When our data set will have a small value data point, the mean will be below the center of data set.

Therefore, the mean of a a set is always affected by outliers and our given statement is false.

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Mrs. Barrera learned that 21 of

otez555 [7]

Answer:

just do 21/25

Step-by-step explanation:

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2 years ago

Find the least common multiple of: 12x^3y^2 and 15x^2y^5

allochka39001 [22]

Answer: $60x^3y^5$

Step-by-step explanation:

Least Common Multiple :The least positive number that is a multiple of two or more numbers.

To find : The least common multiple of: $12x^3y^2$ and $15x^2y^5.$

Since, $12x^3y^2=2\times2\times3\times x^3y^2$

$15x^2y^5=3\times5\times x^2y^5$

The least common multiple of: $12x^3y^2$ and $15x^2y^5.$ = $2\times2\times3\times5\times x^3y^5=60x^3y^5$ [take highest power of x and y ]

Hence, the least common multiple of: $12x^3y^2$ and $15x^2y^5.$ = $60x^3y^5$

3 0

2 years ago

3 1/5 divided by 1 1/7

Katarina [22]

2.8 is the answer for that question

8 0

2 years ago

Read 2 more answers

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of days an

solniwko [45]

Answer:

(a) 283 days

(b) 248 days

Step-by-step explanation:

The complete question is:

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. (a) What is the minimum pregnancy length that can be in the top 11% of pregnancy lengths? (b) What is the maximum pregnancy length that can be in the bottom 5% of pregnancy lengths?

Solution:

The random variable X can be defined as the pregnancy length in days.

Then, from the provided information $X\sim N(\mu=268, \sigma^{2}=12^{2})$ .

(a)

The minimum pregnancy length that can be in the top 11% of pregnancy lengths implies that:

P (X > x) = 0.11

⇒ P (Z > z) = 0.11

⇒ z = 1.23

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\\\1.23=\frac{x-268}{12}\\\\x=268+(12\times 1.23)\\\\x=282.76\\\\x\approx 283$

Thus, the minimum pregnancy length that can be in the top 11% of pregnancy lengths is 283 days.

(b)

The maximum pregnancy length that can be in the bottom 5% of pregnancy lengths implies that:

P (X < x) = 0.05

⇒ P (Z < z) = 0.05

⇒ z = -1.645

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\\\-1.645=\frac{x-268}{12}\\\\x=268-(12\times 1.645)\\\\x=248.26\\\\x\approx 248$

Thus, the maximum pregnancy length that can be in the bottom 5% of pregnancy lengths is 248 days.

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2 years ago

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egoroff_w [7]

Answer:

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Step-by-step explanation:

I hope it's helpful!

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2 years ago