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3 years ago

Question 1

Mathematics

1 answer:

Furkat [3]3 years ago

8 0

Options:

A. Segment EF is twice as long as segment AB.

B. Segment CD is twice as long as segment FG.

C. The measure of angle HEF is twice the measure of angle DAB.

D. The length of segment EH is 16 units.

E. The measure of angle BCD is the same as the measure of angle GFE.

Answer:

A. Segment EF is twice as long as segment AB.

D. The length of segment EH is 16 units.

Step-by-step explanation:

First, let's find out the scale factor that was used in creating quadrilateral EFGH.

Scale factor = ratio of their corresponding segment length.

Segment GH corresponds with segment CD.

Scale factor = GH/CD = 12/6 = 2.

This means all segments that make up quadrilateral ABCD were increased proportionally to give corresponding segment lengths that make up quadrilateral EFGH.

The size of ABCD would increase, while the shape remains the same. This implies that, the measures of corresponding angles are the same, i.e. m<C = m<G.

The following statements are therefore true:

A. Segment EF is twice as long as segment AB.

AB corresponds to EF. Therefore, AB*scale factor = length of EF

AB*2 = EF

EF is twice the length of AB.

D. The length of segment EH is 16 units.

AD corresponds to EH.

AD*scale factor = length of EH

AD*2 = EH

Given that AD = 8, therefore:

8*2 = EH

16 = EH

EH = 16 units

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20×^2-30x-40 solve for x

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Answer:

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Step-by-step explanation:

Solve for x:

360 - 30 x = 0

Subtract 360 from both sides:

(360 - 360) - 30 x = -360

360 - 360 = 0:

-30 x = -360

Divide both sides of -30 x = -360 by -30:

(-30 x)/(-30) = (-360)/(-30)

(-30)/(-30) = 1:

x = (-360)/(-30)

The gcd of 360 and -30 is 30, so (-360)/(-30) = (-(30×12))/(30 (-1)) = 30/30×(-12)/(-1) = (-12)/(-1):

x = (-12)/(-1)

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Answer:

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>

triangles
equation
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<h3>tips and formulas:</h3>

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<h3>let's solve:</h3>

according to the question

2x+2+2x+2+3x+1=180

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Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

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And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

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Part a

The mean can be calculated with this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

Replacing we got:

$\bar X = 369.62$

Part b

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$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And replacing we got:

$s = 621.76$

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