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Mazyrski [523]
4 years ago
12

The lengths of a rectangle have been measured to the nearest tenth of a centimetre they are 87.3cm and 51.8cm what is the upper

bound for the area and the lower bound for the perimeter
Mathematics
2 answers:
vagabundo [1.1K]4 years ago
8 0

Answer: Area (upper bound) = 4527.7056 cm²

               Perimeter (lower bound) = 278 cm

<u>Step-by-step explanation:</u>

The length and width of the rectangle have been ROUNDED to the nearest tenth. Let's calculate what their actual measurements could be:

LENGTH: rounded to 87.3,  actual is between 87.25 and 87.34

<em>87.25 is the lowest number it could be that would round it UP to 87.3</em>

<em>87.34 is the highest number it could be that would round DOWN to 87.3</em>

WIDTH: rounded to 51.8, actual is between 51.75 and 51.84

<em>51.75 is the lowest number it could be that would round it UP to 51.8</em>

<em>51.84 is the highest number it could be that would round DOWN to 51.8</em>

To find the Area of the upper bound, multiply the highest possible length and the highest possible width:

A = 87.34 × 51.84 = 4527.7056

To find the Perimeter of the lower bound, calculate the perimeter using the lowest possible length and the lowest possible width:

P = 2(87.25 + 51.75) = 278

viva [34]4 years ago
4 0

Answer:

Area=4529.0975

Perimeter=278

Step-by-step explanation:

same as the one above but with different area

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Answer:

100% probability that the sample mean scores will be between 87 and 124 points

Step-by-step explanation:

To solve this question, we have 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, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 105, \sigma = 20, n = 20, s = \frac{20}{\sqrt{20}} = 4.47

What is the probability that the sample mean scores will be between 87 and 124 points

This is the pvalue of Z when X = 124 subtracted by the pvalue of Z when X = 87. So

X = 124

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

By the Central Limit Theorem

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

Z = \frac{124 - 105}{4.47}

Z = 4.25

Z = 4.25 has a pvalue of 1

X = 87

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

Z = \frac{87 - 105}{4.47}

Z = -4.25

Z = -4.25 has a pvalue of 0

1 - 0 = 1

100% probability that the sample mean scores will be between 87 and 124 points

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4 years ago
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