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navik [9.2K]
3 years ago
14

Solve for X, assume that all segmented that appear to be tangent are tangent.

Mathematics
1 answer:
Trava [24]3 years ago
8 0

Answer:

Step-by-step explanation:

x = 9

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Please help! I will give 5 star rating and brainlest ! To anyone that helps with this question
kiruha [24]

Answer:

wat is the question?

Step-by-step explanation:

5 0
3 years ago
Chelsey put for $450 into an account with a simple interest rate of 2.5% when she withdrew the money she had earned a total of $
fgiga [73]

Answer:

5.66 years

Step-by-step explanation:

450 x 1.025^(n) = 450 + 67.50

450 x 1.025^(n) = 517.50

1.025^(n) =(517.50 / 450)

1.025^(n) = 1.15

n = ln(1.15) / ln(1.025)

n = 5.66

6 0
3 years ago
The difference of a number and 6 is the same as 5 times the sum of the number and 2. What is the number?
Sunny_sXe [5.5K]

Answer:.OA. -4

Step-by-step explanation:

4 0
3 years ago
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

7 0
2 years ago
Plzz help for brainly
PolarNik [594]

Answer:first one

Step-by-step explanation:

8 0
3 years ago
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