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ikadub [295]
1 year ago
15

A classic counting problem is to determine the number of different ways that the letters of millennium can be arranged. Find tha

t number. The number of different ways that the letters of millennium can be arranged is
Mathematics
1 answer:
Eddi Din [679]1 year ago
7 0

Answer:

<u>The correct answer is that the number of different ways that the letters of the word "millennium" can be arranged is 226,800</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Number of letters of the word "millennium" = 10

Letters repeated:

m = 2 times

i = 2 times

l = 2 times

n = 2 times

2. The number of different ways that the letters of millennium can be arranged is:

We will use the n! or factorial formula, this way:

10!/2! * 2! * 2! * 2!

(10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)/(2 * 1) * (2 * 1) * (2 * 1) * (2 *1)

3'628,800/2*2*2*2 = 3'628,800/16 = 226,800

<u>The correct answer is that the number of different ways that the letters of the word "millennium" can be arranged is 226,800</u>

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A manufacturer produces three models of bicycles. The times (in hours) required for assembling, painting, and packaging each mod
Simora [160]

Answer:

we have to maximize the following equation:

45A + 50B + 55C

where:

A = number of model A bicycles produced

B = number of model B bicycles produced

C = number of model C bicycles produced

the constraints are:

2A + 2.5B + 3C ≤ 4006 (assembly constraint)

A + 0.5B + 2C ≤ 2495 (painting constraint)

A + 0.75B + 1.25C ≤ 1500 (packaging constraint)

A,B,C ≥ 0

using solver, the optimal solution is: 745A + 1006B = $83,825

using slack variables:

2A + 2.5B + 3C + S1 = 4006 (assembly constraint)

A + 0.5B + 2C + S2 = 2495 (painting constraint)

A + 0.75B + 1.25C + S3 = 1500 (packaging constraint)

A,B,C,S ≥ 0

slack variable tableau:

A         B         C           S1          S2          S3           Z            B

2         2.5       3           1             0            0            0            4006

1          0.5       2          0             1            0             0            2495

<u>1          0.75     1.25      0             0           1              0            1500</u>

-45      -50      -55       0             0           0              1            0

4 0
1 year ago
Data show that men between the ages of 20 and 29 in a general population have a mean height of 69.3​ inches, with a standard dev
dangina [55]

Data show that men between the ages of 20 and 29 in a general population have a mean height of 69.3​ inches, with a standard deviation of 2.5 inches. a baseball analyst wonders whether the standard deviation of heights of​ major-league baseball players is less than 2.5 inches. the heights​ (in inches) of  20 randomly selected players are shown in the table.

72 74 71 72 76

70 77 75 72 72

77 72 75 70 73

74 75 73 74 74

What are the correct hypotheses for this  test?

The null hypothesis is H₀?: ____ 2.5

The alternative hypothesis is H₁?: ____  2.5

Calculate the value of the test statistic.

x² = _____ (Round to three decimal places)

Answer:

Null hypothesis, H₀: σ = 2.5

Alternative hypothesis,  Hₐ: μ<2.5

Test statistic = 12.920

Step-by-step explanation:

Given Data shows that:

men between the ages of 20 and 29 in a general population have a mean height of 69.3​ inches, with a standard deviation of 2.5 inches

We consider a random sample of 20 selected baseball players.

Therefore;

The Null and Alternative hypothesis are as follows:

The Null hypothesis is the standard deviation of the heights of major league baseball players is not less than 2.5 inches.

Null hypothesis, H₀: σ = 2.5

On the other hand: The Alternative hypothesis is the standard deviation of the heights of major league baseball players is less than 2.5 inches.  

Alternative hypothesis,  Hₐ: μ<2.5

The Mean Calculation is:

\bar{x} = \frac{1}{2} \sum x_i

= \frac{1}{20} (72+74+...+74) \\ \\ = \frac{1468}{20} \\ \\ =73.4

The sample standard deviation is:

s = \sqrt{\frac{1}{n-1} \sum (x_1 - \bar{x})^2 }

= \sqrt{\frac{1}{20-1} \sum (72-73.4)^2 + ...+(74-73.4)^2 }  \\ \\ =  \sqrt{4.25}  \\ \\ = 2.06

The test statistics is now determined as :

x^2 = \frac{(n-1)s^2}{\sigma^2} \\ \\ = \frac{(20-1)(2.06)^2}{(2.5)^2}  \\ \\ = \frac{19*4.25}{6.25} \\ \\ = \frac{80.75}{6.25} \\ \\ = 12.920

4 0
2 years ago
Please help I don't understand this problem. I need to get this done by tonight.
Lapatulllka [165]

Answer:

Step-by-step explanation:

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4 0
2 years ago
You've run 250 ft of cable that has a loss rate of 3.6 dB per 100 ft. what is your total loss?
Leviafan [203]

Answer:

2.5 dB/100 ft

Explanation:

If 5 dB was lost after 200 ft of cable and 100 ft is half of 200 ft, then the rate of loss should be 2.5 dB per 100 ft.

Step-by-step explanation:

5 0
1 year ago
An office worker can type at the rate of 57 words per minute. Which equation could be used to solve for the number of words he c
8_murik_8 [283]
57x6 =342
57=per min 
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4 0
2 years ago
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