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katrin [286]
3 years ago
9

The Sandberg Institute building in Amsterdam generates revenue by selling advertising space on the exterior of the building. The

building is a rectangular prism with dimensions 50 m by 40 m by 75 m. Suppose it costs 1 Euro per month to rent an advertising space of 50 cm2 . Each of the 4 walls of the building is covered with advertisements. How much money will the institute earn in one month?
Mathematics
1 answer:
Contact [7]3 years ago
5 0

Answer:

2.7 million euros per month

Step-by-step explanation:

Since the room is a rectangular prism, we find the area of the room without its top and bottom, which is the area of the walls, A. So,

A = 2(lh + wh) = 2h(l + w) where l = length of room = 50 m, w = width of room = 40 m and h = height of room = 75 m

Substituting the values of the variables into A, we have

A = 2h(l + w)

= 2(75 m)(50 m + 40 m)

= 150 m × 90 m

= 13500 m²

= 13500 × 1m²

= 13500 × 10000 cm²

= 135000000 cm²

= 1.35 × 10⁸ cm²

Now, since 50 cm² of space = 1 Euro per month, then 1.35 × 10⁸ cm²

= 1.35 × 10⁸ cm² × 1 Euro per month/50 cm²

= 0.027 × 10⁸ Euro per month

= 2.7 × 10⁶ Euro per month

= 2.7 million euros per month

So, the institute will earn 2.7 million euros per month

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Answer:

A customer who sends 78 messages per day would be at 99.38th percentile.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Average of 48 texts per day with a standard deviation of 12.

This means that \mu = 48, \sigma = 12

a. A customer who sends 78 messages per day would correspond to what percentile?

The percentile is the p-value of Z when X = 78. So

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

Z = \frac{78 - 48}{12}

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0.9938*100% = 99.38%.

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