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kotykmax [81]
2 years ago
15

Nadia’s bookshelf contains 10 fiction books, two reference books, and five nonfiction books. What is the probability that she ra

ndomly picks up a reference book and then, without replacing it, picks up a nonfiction book?
Mathematics
2 answers:
pshichka [43]2 years ago
7 0

Answer:

5/136

Step-by-step explanation:

Probability is the chance of an event happening.

Probability = (Favorable outcome)/(total outcome)

P(r and n) = 2/17 × 5/16

=(2 × 5)/(17 × 16)

= 10/272

= 5/136

lawyer [7]2 years ago
4 0

Answer:

The required probability is \frac{5}{136}.

Step-by-step explanation:

Consider the provided information.

It is given that the total number of reference book is 2.

Fiction books are 10 and non fiction books are 5.

Thus, the total book are: 5 + 10 + 2 = 17

The probability that Nadia picks up a reference book is:

P(Reference) = \frac{2}{17}

Now with out replacing it she pick up another book.

The probability that she randomly picks up a non-fiction book is:

P(Nonfiction) = \frac{5}{16}

Now we need to find the probability that she randomly picks up a reference book and then, without replacing it, picks up a nonfiction book.

In order to find the required probability multiply both the probability as shown:

P(\text{Nonfiction}|\text{Reference})=\frac{2}{17}\times\frac{5}{16}

P(\text{Nonfiction}|\text{Reference})=\frac{5}{136}

Hence, the required probability is \frac{5}{136}.

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A charity fair raised $6,000 by selling 500 lottery tickets. There were two types of lottery tickets: A tickets cost $10 each, a
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4107 cells

Step-by-step explanation:

From the question, we have the following values:

Day 1 : 113 cells

Number of cells increases by day by 82%

Hence,

Day 2

113 × 82% = 92.66cells

Hence, Total number of bacteria cells for Day 2 = 113 + 92.66 = 205.66cells

Day 3

205.66 × 82% = 168.6412 cells

Hence, Total number of bacteria cells for Day 3 = 168.6412 + 205.66 = 374.3012 cells

Day 4

374.3012 × 82% = 306.926984 cells

Hence, Total number of bacteria cells for Day 4 = 306.926984 + 374.3012 = 681.228184 cells

Day 5

681.228184 × 82% = 558.60711088 cells

Hence, Total number of bacteria cells for Day 5 = 558.60711088 + 681.228184 = 1239.8352949 cells

Day 6

1239.8352949 × 82% = 1016.6649418 cells

Hence, Total number of bacteria cells for Day 5 = 1016.6649418 + 1239.8352949 = 2256.5002367 cells

Day 7

2256.5002367 × 82% = 1850.3301941 cells

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Approximately to nearest whole number, the total number of bacteria cells that would be present after 7 days = 4107 cells

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