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Finger [1]
3 years ago
10

There are 318 fiction books in the class library. The number of nonfiction books is 47 less than the number of fiction books. Ho

w many fiction and non fiction books are there in the class library
Mathematics
2 answers:
grigory [225]3 years ago
6 0

Answer:

589 books in total

Step-by-step explanation:

fiction books= 381

non-fiction= 381-47= 271

total amount of books(ficton books and non-fiction books together)= 589 books in total

MissTica3 years ago
6 0

Answer:

589 Books

Why?

Well, as you can see, there are 318 <em>fiction</em><em> </em>books in the library. You don't yet know the number of <em>nonfiction</em> books, so let's just focus on that right now.

<u><em>Important</em><em> </em><em>Information</em><em> </em><em>To</em><em> </em><em>Remember</em><em> </em></u>

Number of fiction books- 318

Number of nonfiction books- 47 less than number of fiction books

<u>Nonfiction</u><u> </u>

The problem states that the number of nonfiction books is 47 less than the number of fiction books. Well, you know that there are 318 fiction books. The number of nonfiction books if 47 <em>less</em><em> </em>than the number of fiction books. Therefore, the equation to solve the number of nonfiction books is 318-47. That is equal to 271.

<u>Nonfiction</u><u> </u><em><u>and</u></em><em><u> </u></em><u>Fiction</u><u> </u>

Well, now we know the number of fiction (318) <em>and</em><em> </em>nonfiction (271). Now, read the last question. "How many fiction <em>and</em> non fiction books are there in the class library?" Well, here, you just simply add the number of fiction and nonfiction together. 318+271. This is equal to 589. Therefore, there are 589 fiction and nonfiction books in the class library.

<em>I</em><em> </em><em>hope</em><em> </em><em>I</em><em> </em><em>helped</em><em>!</em><em> </em><em>Let</em><em> </em><em>me</em><em> </em><em>know</em><em> </em><em>if</em><em> </em><em>you</em><em> </em><em>have</em><em> </em><em>any</em><em> </em><em>questions</em><em>!</em><em> </em><em>:</em><em>)</em>

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A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20
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a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that p = 0.7

20 adult women

This means that n = 20

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

P(X < 14) = 1 - P(X \geq 14)

In which

P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916

P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789

P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304

P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716

P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278

P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068

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P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079

P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921

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(b) At least 17 of them have had a physical examination in the past two years

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P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

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