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Usimov [2.4K]
3 years ago
15

I am greater than 4 tens and less than 5 tens I have 9 ones

Mathematics
1 answer:
aliina [53]3 years ago
4 0

4 tens = 40 and 5 tens = 50 … therefore the number has to be between 40 and 50

So: 40 < x <50 … this says that x is greater than 40 and less than 50

 

So the possible numbers would be: 41,42,43,44,45,46,47,48,49

Now the number must have 9 ones … which mean the only number would be 49

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you measure the period of a mass oscillating on a vertical spring ten times as follows: period (s): 1.06, 1.31, 1.28, 0.99, 1.48
lidiya [134]

The mean and (sample) standard deviation σ = 0.2098.

<h3>What exactly would the standard deviation indicate?</h3>

The term "standard deviation" (or "") refers to the degree of dispersion of the data from the mean. Data are grouped around the mean when the standard deviation is low, and are more dispersed when the standard deviation is high.

<h3>According to given information:</h3>

The mean is the product of the dataset's total and the sample size. Mathematically.

\bar{x}=\frac{\sum X_i}{N}

The individual periods are Xi.

The sample size is N.

\sum X i = 1.06 + 1.31 + 1.28 + 0.99,+  1.48 + 1.37+  0.98 + 1.31 + 1.59 + 1.55

\sum X i = 12.92

N = 10

While substituting the value we get:

x = 12.96/10

x = 1.292

The samples' average is 1.292.

The standard deviation:

\sigma=\sqrt{\frac{\sum(x-\bar{x})^2}{N}}

\sum(x-\bar{x})^2 = (1.48-1.292)^2+(1.37-1.292)^2+(0.98-1.292)^2+(1.31-1.292)^2+(1.59-1.292)^2+(1.55-1.292)^2.

\sum(x-\bar{x})^2 = 0.43996

Putting into the formula we get:

\sigma=\sqrt{\frac{0.43996}{10}}

σ = √(0.043996)

σ = 0.2098

The mean and (sample) standard deviation σ = 0.2098.

To know more about standard deviation visit:

brainly.com/question/18521100

#SPJ4

I understand that the question you are looking for is:

You measure the period of a mass oscillating on a vertical spring ten times as follows:

Period (s): 1.06, 1.31, 1.28, 0.99, 1.48, 1.37, 0.98, 1.31, 1.59, 1.55

Required:

What are the mean and (sample) standard deviation?

a. Mean: 1.228, Standard Deviation: 0.2135

b. Mean: 1.325, Standard Deviation: 0.1674

c. Mean: 1.292. Standard Deviation: 0.2211

d. Mean: 1.228, Standard Deviation: 0.2098

e. Mean: 1.292, Standard Deviation: 0.2135

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1 year ago
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