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Rudiy27
3 years ago
7

WILL GIVE A BRAINLIEST TO THE CORRECT ANSWER PLEASE HELP!!

Mathematics
1 answer:
ivolga24 [154]3 years ago
4 0
Hey

km/3x-2x= 4

km/3=4+2x

km=3(4+2x)

km=12+6x

km-12=6x

x=km-12/6


so D is the answer

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Seventeen is three times the difference between four times a number and five
expeople1 [14]

17 = 3(4x - 5)

<em><u>Distributive property.</u></em>

17 = 12x - 15

<em><u>Add 15 to both sides.</u></em>

32 = 12x

<em><u>Divide both sides by 12.</u></em>

x = 2.67. (This is your answer.)

Let me know if you have any questions.

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Given the functions k(x) = 5x − 8 and p(x) = x − 4, solve k[p(x)] and select the correct answer below.
Mazyrski [523]
P(x) = x-4

k(x-4) = 5(x-4) - 8 = 5x - 28

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3 years ago
The sum of two numbers is 64 and the difference is 6. What are the numbers?
Ne4ueva [31]

Answer:

29 and 35

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x+y=64

x-y=6

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If angles are congruent, solve by setting expressions equal to each other.<br><br> True<br> False
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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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