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olchik [2.2K]
3 years ago
6

a bookshelf holds 43 books on each shelf is the total number of books proportional to the number of shelves in the bookshelf

Mathematics
2 answers:
Mama L [17]3 years ago
7 0
If every shelf has the same amount of books, then it can be seen as proportional because one shelf has 43 and 2 have 86, and it keeps going up in a consistent way. 
so 43/1 = 86/2 .
Happy studying ^-^
pochemuha3 years ago
5 0

Answer:

Yes, the total number of books is proportional to the number of shelves in the bookshelf.

Step-by-step explanation:

Let y represent total number of books and x represent number of shelves.

We have been given that a bookshelf holds 43 books on each shelf. We are asked to determine whether the total number of books proportional to the number of shelves in the bookshelf.

The total number of books on x shelves would be 43x. We can represent this information in an equation as:

y=43x

Therefore, the total number of books is proportional to number of shelves in the bookshelf and constant of proportionality is 43.

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Adam cut a piece of string 42 inches long into 2 pieces. The ratio of the length of the 2 peices is 3:4, what is the length, to
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Answer:

Shorter piece length is 18 inches

Step-by-step explanation:

Let's marked first piece with x and second with y

x + y = 42

Given ratio is x : y = 3 : 4

From this proportion we get

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when we replace this in initial equation we get

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6 0
2 years ago
A cookie factory monitored the number of broken cookies per pack yesterday.
trapecia [35]

Answer:

Confidence Interval - 2.290 < S < 2.965

Step-by-step explanation:

Complete question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Solution

Given  

n=50

x=23.33

s=2.6

Alpha = 1-0.80 = 0.20  

X^2(a/2,n-1) = X^2(0.10, 49) = 63.17

sqrt(63.17) = 7.948

X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69

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s*sqrt(n-1) = 18.2

s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } } \leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }

confidence interval:

(18.2/7.948) < S < (18.2/6.139)

2.290 < S < 2.965

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