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enot [183]
2 years ago
13

How many books can be stacked on top of one another inside a cupboard 1.21 m high on the inside, knowing that the thickness of e

ach book is 12.5 cm.
Mathematics
1 answer:
kvasek [131]2 years ago
4 0

Answer:

10 books approx

Step-by-step explanation:

Step one:

Given data

Hight of cupboard= 1.21m

Thickness of book= 12.5cm

thickness in meters= 12.5/100= 0.125m

Required

The number of books need to fill the Cupboard

Step two:

the number of books can be computed as

= 1.21/0.125

=9.68 book

=10 books approx

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Find pb, if pj=14 and jb= 28
anyanavicka [17]

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42

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3 years ago
Read 2 more answers
Solve the problem. Use the Central Limit Theorem.The annual precipitation amounts in a certain mountain range are normally distr
bazaltina [42]

Answer:

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

Step-by-step explanation:

To solve this question, we use the normal probability distribution and the central limit theorem.

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.

Central Limit Theorem

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 109.0 inches, and a standard deviation of 12 inches.

This means that \mu = 109, \sigma = 12

Sample of 25.

This means that n = 25, s = \frac{12}{\sqrt{25}} = 2.4

What is the probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches?

This is the p-value of Z when X = 112. So

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

By the Central Limit Theorem

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

Z = \frac{112 - 109}{2.4}

Z = 1.25

Z = 1.25 has a p-value of 0.8944.

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

7 0
3 years ago
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