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NemiM
2 months ago
9

Emi computes the mean and variance for the population data set 87, 46, 90, 78, and 89. She finds the mean is 78. Her steps for f

inding the variance are shown below. mc008-1.jpg What is the first error she made in computing the variance? Emi failed to find the difference of 89 - 78 correctly. Emi divided by N - 1 instead of N. Emi evaluated (46 - 78)2 as -(32)2. Emi forgot to take the square root of -135.6.
Mathematics
2 answers:
AnnZ [12.3K]2 months ago
7 0
Let’s determine the actual mean

First, we sum all the values

87+46+90+78+89 = 390

Then, we divide 390 by the count of numbers present.

390/5 = 78

Thus, the mean is 78

Emi did not manage to calculate the difference
tester [12.3K]2 months ago
4 0
Correct answer is C. <span>Emi evaluated (46 - 78)2 as -(32)2.</span>
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A jewelry artist is selling necklaces at an art fair. It costs $135 to rent a booth at the fair. The cost of materials for each
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The artist begins to profit when n > 8.18

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Step-by-step explanation:

Given

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8 0
2 months ago
A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm, and 96 cm but does not resonate at any wave
Zina [12379]

Response:

A 120 cm long pipe produces resonance at wavelengths of 480 cm, 160 cm, and 96 cm, but fails to resonate with any wavelengths longer than those. Consequently, this pipe is classified as:

A. closed at both ends

B. open at one end and closed on the other

C. open at both ends.

D. we cannot determine since we lack information about the sound frequency.

The correct answer is:

B. open at one end and closed at the other.

Explanation in steps:

The given data states that the pipe has a length of

= 120 cm

with wavelengthsL = 480 cm

                             

= 160 cm and \lambda_1 = 96 cm

We need to ascertain if the pipe is open, closed, or of an open-closed nature.

\lambda_2Note:\lambda_3

For a pipe open at both ends, the fundamental wavelength is 2L.

For a pipe closed at one end and open at the other, the fundamental wavelength is 4L.

  • From this information,
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Now let’s verify with the subsequent wavelengths.

In the case of a pipe with one open end and one closed end:[

Odd multiples of one-quarter wavelength must fit within the length L4L=4(120)=480\ cm.

⇒  

                           ⇒  

⇒

                               ⇒  

⇒                                    ⇒  ⇒

                             ⇒  

 Thus, it confirms that the pipe is open at one end and closed at the other.

6 0
1 month ago
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