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Fofino [41]
3 years ago
11

A random sample of 100 workers in one large plant took an average of 12 minutes to complete a task, with a standard deviation of

2 minutes. A random sample of 50 workers in a second large plant took an average of 11 minutes to complete the task, with a standard deviation of 3 minutes. Construct a 95% confidence interval for the difference between the two population mean completion times.
Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
5 0

Answer:

The 95% Confidence Interval for the difference between the two population mean completion times =

(0.081, 1.919)

Step-by-step explanation:

Confidence Interval for difference between two means =

μ1 -μ2 ± z × √ σ²1/n1 + σ²2/n2

Where

μ1 = mean 1 = 12 mins

σ1 = Standard deviation 1 = 2 mins

n1 = 100

μ2= mean 2 = 11 mins

σ2 = Standard deviation 2 = 3 mins

n1 = 50

z score for 95% confidence interval = 1.96

μ1 -μ2 ± z × √ σ²1/n1 + σ²2/n2

= 12 - 11 ± 1.96 × √2²/100 + 3²/50

= 1 ± 1.96 × √4/100 + 9/50

= 1 ± 1.96 × √0.04 + 0.18

= 1 ± 1.96 × √0.22

= 1 ± 1.96 × 0.469041576

= 1 ± 0.9193214889

Confidence Interval

= 1 - 0.9193214889

= 0.0806785111

≈ 0.081

1 + 0.9193214889

= 1.9193214889

≈ 1.919

Therefore, the 95% Confidence Interval for the difference between the two population mean completion times =

(0.081, 1.919)

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olya-2409 [2.1K]

Answer:

<em>DF = 10 units</em>

m\angle DFG = 28^\circ

<em>EG = 5.04</em>

Step-by-step explanation:

<u>Properties of Rhombus es</u>

  1. All Sides Of The Rhombus Are Equal.
  2. The Opposite Sides Of A Rhombus Are Parallel.
  3. Opposite Angles Of A Rhombus Are Equal.
  4. In A Rhombus, Diagonals Bisect Each Other At Right Angles.
  5. Diagonals Bisect The Angles Of A Rhombus.

The image contains a rhombus with the following data (assume the center as point O):

DO = 5 units

GF = 5.6 units

m\angle FEO = 62^\circ

4. Calculate DF

Applying property 4, diagonals bisect each other, thus the length of DF is double the length of DO, i.e. DF=2*5 = 10:

DF = 10 units

5. Calculate m\angle DFG

Applying property 4 in triangle EFO, the center angle is 90°, thus angle EFO has a measure of 90°-62°=28°.

Applying property 5, this angle is half of the measure of angle EFG and angle DFG has the same measure of 28°.

m\angle DFG = 28^\circ

6. FG is the hypotenuse of triangle OFG, thus:

OG^2=FG^2-OF^2

OG^2=5.6^2-5^2

OG^2=6.36

OG=\sqrt{6.36}=2.52

EG is double OG: OG=2*2.52=5.04

EG = 5.04

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2 years ago
The altitude drawn to the hypotenuse of a right triangle divides the hypotenuse into segments such that their lengths are in rat
Zanzabum

In geometry, it would be always helpful to draw a diagram to illustrate the given problem.

This will also help to identify solutions, or discover missing information.

A figure is drawn for right triangle ABC, right-angled at B.

The altitude is drawn from the right-angled vertex B to the hypotenuse AC, dividing AC into two segments of length x and 4x.


We will be using the first two of the three metric relations of right triangles.

(1) BC^2=CD*CA (similarly, AB^2=AD*AC)

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Part (A)

From relation (2), we know that

BD^2=CD*DA

substitute values

8^2=x*(4x) => 4x^2=64, x^2=16, x=4

so CD=4, DA=4*4=16 (and AC=16+4=20)


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Using relation (1)

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again, substitute values

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=>

AB

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