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lubasha [3.4K]
3 years ago
9

PLEASE HELP. I WILL VERY MUCH APPRECIATE IT!!!

Mathematics
1 answer:
alexgriva [62]3 years ago
3 0

Answer:

10 centimeters

Step-by-step explanation:

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You don't have to do them all
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Hope this helps an mark brainliest please!

3 0
2 years ago
There are 10 employees in a particular division of a company. Their salaries have a mean of 570,000, a median of $55,000,and a s
harkovskaia [24]

Answer:

a) $160,000

b) $55,000

c) $332264.804

Step-by-step explanation:

We are given that there are 10 employees in a particular division of a company and their salaries have a mean of $70,000, a median of $55,000, and a standard deviation of $20,000.

And also the largest number on the list is $100,000 but By accident, this number is changed to $1,000,000.

a) Value of mean after the change in value is given by;

     Original Mean = $70,000

       \frac{\sum X}{n} = $70,000  ⇒ \sum X = 70,000 * 10 = $700,000

   New \sum X after change = $700,000 - $100,000 + $1,000,000 = $1600000

  Therefore, New mean = \frac{1600000}{10} = $160,000 .

b) Median will not get affected as median is the middle most value in the data set and since $1,000,000 is considered to be an outlier so median remain unchanged at $55,000 .

c) Original Variance = 20000^{2} i.e.  20000^{2} = \frac{\sum x^{2} - n*xbar }{n -1}

    Original \sum x^{2} = (20000^{2} * (10-1)) + (10 * 70,000) = $3,600,700,000

    New \sum x^{2} = $3,600,700,000 - 100,000^{2} + 1,000,000^{2} = 9.936007 * 10^{11}  

    New Variance = \frac{new\sum x^{2} - n*new xbar }{n -1} = \frac{9.936007 *10^{11}  - 10*160000 }{10 -1} = 1.103999 * 10^{11}    Therefore, standard deviation after change = \sqrt{1.103999 * 10^{11} } = $332264.804 .

7 0
3 years ago
I don’t know how to do it, please teach me !!!
cricket20 [7]

As described in z-distribution the answers are given below:

a) The 95% confidence interval estimate for the population mean spending by a customer on visiting salon per month is given as follows: (747, 853).

b) The sampling error at 95% confidence level is of: $35.78.

What is a z-distribution ?

The normal distribution with a mean of 0 and a standard deviation of 1 is referred to as the standard normal distribution (also known as the Z distribution) (the green curves in the plots to the right). It is frequently referred to as the bell curve since the probability density graph resembles a bell.

solution:

The bounds of the confidence interval are given as follows:

In which:

is the sample mean.

z is the critical value.

n is the sample size. is the standard deviation for the population.

The parameters for this problem are given as follows:

Hence the lower bound of the interval is of:

800 - 200 x 1.96/square root of 55  = 747.

The upper bound of the interval is of:

800 + 200 x 1.96/square root of 55  = 853.

The sampling error for a sample size of 120 is calculated as follows:

200 x 1.96/square root of 120 = $35.78.

To learn more about the z-distribution from the given link

brainly.com/question/4079902

#SPJ1

5 0
1 year ago
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