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4 years ago

What is the name of the other main library classification system besides the dewey decimal system ?

Mathematics

2 answers:

Strike441 [17]4 years ago

6 0

Answer:

Online Public Access catalog.

Explanation:

Library classification system help libraries to manage and store thousands of book and materials, it is the system which is developed to keep the library resources arranged in systematical order.

Library manages thousands of material of different subjects that need to settle in a manner that allows the easiest possible access to the end-user and he does not need to face any problem.

Assoli18 [71]4 years ago

4 0

OPAC - Online Public Access Catalog

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1. Which is greater than 4? (a) 5, (b) -5, (c) -1/2, (d) -25.

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Answer:

(a) 5

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3 years ago

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A stack of 30 science flashcards includes a review card for each of the following: 10 insects, 8 trees, 8 flowers, and 4 birds.

Gnom [1K]

The probability of randomly selecting a tree or a flower is 8/15.

<h3>What is the probability?</h3>

Probability determines the chances that an event would happen. The probability the event occurs is 1 and the probability that the event does not occur is 0.

The probability of randomly selecting a tree or a flower = (number of trees / total number of flash cards) + (number of flowers / total number of flash cards)

8/30 + 8/30 = 16/30 = 8/15

To learn more about probability, please check: brainly.com/question/13234031

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2 years ago

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3 years ago

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zavuch27 [327]

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3 years ago

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A professional basketball team made 37.9% of its three-point field goals in one season. If 80 three-point field goal attempts ar

kompoz [17]

Answer:

The probability that more than 35 were made is 0.14686.

Step-by-step explanation:

We are given that a professional basketball team made 37.9% of its three-point field goals in one season.

80 three-point field goal attempts are randomly selected from the season.

Let $\hat p$ = sample proportion of three-point field goals made in one season.

The z-score probability distribution for the sample proportion is given by;

Z = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }$ ~ N(0,1)

where, p = population proprotion of three-point field goals = 37.9% = 0.379

n = sample of three-point field goal attempts = 80

$\hat p$ = sample proportion of three-point field goals = $\frac{35}{80}$ = 0.4375

Now, if 80 three-point field goal attempts are randomly selected from the season, the probability that more than 35 were made is given by = P( $\hat p$ > 0.4375)

P( $\hat p$ > 0.4375) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }$ > $\frac{0.4375-0.379}{\sqrt{\frac{0.4375(1-0.4375)}{80}} }$ ) = P(Z > 1.05) = 1 - P(Z $\leq$ 1.05)

= 1 - 0.85314 = 0.14686

The above probability is calculated by looking at the value of x = 1.05 in the z table which has an area of 0.85314.

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