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frozen [14]
2 years ago
6

As an event organizer, you are to host 120guests in a room of dimensions 18m long and 12m wide. You have rectangular tables whic

h are 2m long and 0.75m wide. The number of guests allocated for the top table is 12 and the rest of the guests are on 3springs. NB: cover space is 0.6m i) Sketch the table plan arrangement. ii) Calculate the number of tables required for the top table. iii) Calculate the number of covers on each spring. iv) calculate the number of tables required for each spring. v) What is the purpose of leaving a gangway when arranging tables and chairs for an event?
Mathematics
2 answers:
Free_Kalibri [48]2 years ago
6 0
Answer:
length of the table=2m cover space for a guests=0.6m so cover space for top table is 12=12×0•6=7•3(4×2)=8



Step-by-step explanation:

crimeas [40]2 years ago
3 0

The room, the tables and the guests are illustrations of rectangles and areas.

<h3>How many tables are required for the top table?</h3>

This is calculated using:

Tables = Guest ÷ Guest per top table

So, we have:

Tables = 120 ÷ 12

Evaluate

Tables = 10

Hence, 10 tables are required for the top table

<h3>How many covers are on each spring?</h3>

This is calculated using:

Covers = Number of spring ÷ Cover space

So, we have:

Covers = 3 ÷ 0.6

Evaluate

Covers = 5

Hence, 5 covers are required for each spring

<h3>How many tables are required for each spring?</h3>

This is calculated using:

Tables = Tables for top table ÷ Spring

So, we have:

Tables = 10 ÷ 3

Evaluate

Tables = 3.33

Remove decimal

Tables = 3

Hence, 3 tables are required for each spring

<h3>What is the purpose of leaving a gangway when arranging tables and chairs for an event?</h3>

This is done to allow passage for walkway

Read more about areas at:

brainly.com/question/24487155

#SPJ2

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Answer:

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.63

If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 months

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{120.1 - 119}{1.63}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

Z = \frac{X - \mu}{s}

Z = \frac{117.9 - 119}{1.63}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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