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Elis [28]
3 years ago
14

Four students are asked to draw a parallelgram. each drew a differently-named figure

Mathematics
1 answer:
nikitadnepr [17]3 years ago
3 0
So basicly the question is asking, name 4 parrelegrams. Thr answer will be, Rectangle, Squrare Trapizoid, and Rhymbus. Not is a specifc order.
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Please help.i am really struggling
alina1380 [7]

Answer:

3,-3) becomes ; (3 + 5 , -3-12) ; (8,-15)

(7,-10) becomes;( 7 + 5, -10-12) ; (12,-22)

(13,-14) becomes (7 + 13, -14-10) ; (20,-24)

Step-by-step explanation:

What we have to do here is to add 5 to the x-axis value and subtract 12 from the y-axis value

(3,-3) becomes ; (3 + 5 , -3-12) ; (8,-15)

(7,-10) becomes;( 7 + 5, -10-12) ; (12,-22)

(13,-14) becomes (7 + 13, -14-10) ; (20,-24)

3 0
3 years ago
Hgunginwn v snvsdngsnanfanvsnsnshnsknbsjnjspejgnjdbhkjdkjf<br><br><br><br> whats sleep
Rasek [7]

Answer:

Step-by-step explanation:

when you lay down and close your eyes

7 0
3 years ago
Read 2 more answers
Find the difference <br><br><br>-2(c + 2.5) - 5(1.2c + 4)​
Vikentia [17]
Answer:
|−∣=7
Explanation:

∣−∣
= |(-2-5)|
= |-2-5|
= | -7|=7
8 0
3 years ago
Suppose a railroad rail is 4 kilometers and it expands on a hot day by 15 centimeters in length. Approximately how many meters w
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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
2 years ago
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