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Eddi Din [679]
3 years ago
10

PLEASE HELP Graph the ordered pairs for y = 3x + 3 using r = {-2, 1, 2}.

Mathematics
2 answers:
rosijanka [135]3 years ago
8 0

Answer:

yes

Step-by-step explanation:

Fed [463]3 years ago
5 0

9514 1404 393

Answer:

  (-2, -3), (1, 6), (2, 9) are plotted in the attached graph

Step-by-step explanation:

For x = -2, y = 3(-2) +3 = -3. The ordered pair is (-2, -3).

For x = 1, y = 3(1) +3 = 6. The ordered pair is (1, 6).

For x = 2, y = 3(2) +3 = 9. The ordered pair is (2, 9).

The graph is attached.

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Multiply. - 4.3 x - 3.2 = _____.
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Help me to do this please
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If x represents the mass of 1 ball in kg, then the balance shows ...

... 6x = x + 9 kg

... 5x = 9 kg

Then 6x is ...

... (6/5)·5x = (6/5)·9 kg = 10.8 kg

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so 10.8 kg is ...

... D. 10 kg 800 g

7 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
1 year ago
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