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olchik [2.2K]
3 years ago
11

A manager at the bookstore must arrange a collection of 231 books for display. The manager decides to use a combination of 15 sh

elves and tables to display the books. He places 7 books on each shelf and 25 books on each table. How many tables did he use for the display?
Mathematics
1 answer:
Kitty [74]3 years ago
6 0

Answer:

6 tables.

Step-by-step explanation:

If he has 15 shelves, we can put the overall number of books off to the side for a moment. First calculate how many books he COULD put onto shelves. This would be 7*15, since each of fifteen shelves can hold 7 books. Counting by fifteens or using a calculator allows us to see that the shelves can hold 105 books.

Going back to the original number of books, we subtract 105 from the original value. The equation at this point is 231 - 105. The result is 126 books left to put on top of tables. But we're not done yet!

Since a table can hold 25 books, we need to divide the remaining number of books by 25. That would be 126/25. Doing this gives us 5 tables we would need, plus one book let over. Since we've run out of shelves, we MUST use another table just for the final book. That's 5+1, or 6 tables. Let's not forget to label our answer.

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Musya8 [376]

Answer:

78

Step-by-step explanation:

Given that:

First day (a) = 1

1 more pebble every subsequent day than the previous day

Tn = a + (n - 1)d

d = common difference ; difference between pebbles in two successive days = 1

n = nth day

At the end of the 12th day;

Tn = a + (n - 1)d

T(12) = 1 + (12 - 1) 1

T(12) = 1 + 11

T(12) = 12

Appling the sun if arithmetic progression formula : Sum of AP:

n/2 (a + Tn)

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12/2 (1 + 12)

6(13)

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4 0
3 years ago
In a statewide lottery, one can buy a ticket for $1. With probability .0000001, one wins a million dollars ($1,000,000), and wit
Dahasolnce [82]

Using the expected value, it is found that the mean of the distribution equals $0.1.

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The probabilities of <u>each outcome</u> are:

  • .0000001 probability of earning $1,000,000.
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Thus, the mean is given by:

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Thus showing that the expected value is $0.1.

A similar problem is given at brainly.com/question/24855677

6 0
2 years ago
If there were 49 cars in a line that stretched 528 feet, what is the average car length? assume that the cars are lined up bumpe
lakkis [162]

If there were 49 cars in a line that stretched 528 feet, what is the average car length? assume that the cars are lined up bumper-to-bumper

Answer: We are given there are 49 cars

Also these 49 cars are in a line stretched 528 feet.

Now the average length of the car is:

Average length = \frac{(length-covered-by-49-cars)}{Number-of-cars}

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Therefore, the average length of car is 10.78 feet

                 

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3 years ago
Helpppppp mehhhhh (^0^)
Norma-Jean [14]
1. Ten
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3 0
2 years ago
Find the volume of a cone with a radius of 8 cm and height 15 cm? Round to the nearest tenth. Please show work.
Nonamiya [84]
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