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tigry1 [53]
3 years ago
10

If bcd~gef find value of x

Mathematics
1 answer:
ddd [48]3 years ago
3 0

Answer:

x = 13

Step-by-step explanation:

Given

See attachment

Required

Find x

From the attachment, we have:

BC = x + 4

BD = 2x - 7

GE = 51

GF = 57

Because both triangles are similar, we use the following equivalent ratios:

BC : BD = GE : GF

Substitute values for each parameter

x + 4 : 2x- 7 = 51 : 57

Convert to fractions

\frac{x + 4}{2x- 7} = \frac{51 }{ 57}

Simplify the right-hand side

\frac{x + 4}{2x- 7} = \frac{17}{19}

Cross Multiply

19(x + 4) = 17(2x - 7)

19x + 76 = 34x - 119

Collect Like Terms

19x - 34x = -76 - 119

-15x = -195

Solve for x

x = \frac{-195}{-15}

x = 13

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Step-by-step explanation:

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Tiembra bought a bag of 36 apples. She used 25% of the apples to make an apple cake and 1/3 of the apples to make an apple pie.
sattari [20]
36×0.25=9 and 9÷3=6. So 6 apples are left after making these desserts.
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2x + 8y = 6<br> -5x - 20y = -15
kodGreya [7K]

Answer:

The total would be 40.53

Or

Solve for the first variable in one of the equations, then substitute the result into the other equation.

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Step-by-step explanation:

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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
2 years ago
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