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Pie
3 years ago
6

Marla writes the problem, 628 divided by 1300.

Mathematics
2 answers:
Maslowich3 years ago
8 0

Answer:

628/1300

Step-by-step explanation:

These are what would represent her problem..

Lelechka [254]3 years ago
6 0

Answer:

628 over 1300

628/1300

I really hope this helps!

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A model shows the visualize of what you need to do in order to complete the problem your asked. It can also show important steps and make it easier for you to answer the question, like length×height×width.
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3 years ago
A sample of 50 observations is taken from an infinite population. The sampling distribution of : a.is approximately normal becau
Korvikt [17]

Answer:

a.is approximately normal because of the central limit theorem.

Step-by-step explanation:

The central limit theorem states that if we have a population with mean μ and standard deviation σ and we take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed.

For any distribution if the number of samples n ≥ 30, the sample distribution will be approximately normal.

Since in our question, the sample of observations is 50, n = 50.

Since 50 > 30, then <u>our sample distribution will be approximately normal because of the central limit theorem.</u>

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3 years ago
The following probability distributions of job satisfaction scores for a sample of information systems (IS) senior executives an
Mashcka [7]

Answer:

Step-by-step explanation:

To calculate ;

1) the expected value of the job satisfaction score for senior executives ;

expected value = Summation (Px)

= 1 x 0.05 + 2 x 0.09 + 3 x 0.03 + 4 x 0.42 + 5 x 0.41

= 4.05

2) the expected value of the job satisfaction score for middle managers;

= 1 x 0.04 + 2 x 0.10 + 3 x 0.12 + 4 x 0.46 + 5 x 0.28

= 3.84

c) the variance of job satisfaction scores for executives and middle managers (to 2 decimals).

Executives ; Variance = Summation(PX^2 - Summation(PX)^2

i) For Executive Managers = 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2 = 1.246 = 1.25

ii) for middle managers ; 1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 = 1.134 = 1.13

d) the standard deviation of job satisfaction scores for both probability distributions (to 2 decimals). Executives, Middle managers;

For Executives = square root [ 1 x 0.05 + 2^2 x 0.09 + 3^2 x 0.03 + 4^2 x 0.42 + 5^2 x 0.41 - 4.05^2] = 1.12

For Middle Managers ; Square root [1 x 0.04 + 2^2 x 0.10 + 3^2 x 0.12 + 4^2 x 0.46 + 5^2 x 0.28 - 3.84^2 ] = 1.06

e) from the values gotten for the variance of both executive and middle managers, the variance of the former is more than that of the latter as such higher satisfaction with the executive managers.

5 0
3 years ago
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4

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