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Brums [2.3K]
3 years ago
8

Suppose that 14 inches of wire costs 84 cents. At the same rate, how many inches of wire can be bought for 48 cents

Mathematics
2 answers:
Valentin [98]3 years ago
8 0
We can set up a simple ratio to solve this.

14 inches/84 cents = x inches / 48

14/84 = x/48

(14/84)*48 = (x/48)*48

(14/84)*48 = x   (the 48's on the right side cancel out)

14*48 / 84 = x

672/84 = x

8 = x

You can buy 8 inches of wire with 48 cents.


Reason this works: The price is the same per inch, so both ratios should equal each other. (if you don't understand this let me know)


AleksandrR [38]3 years ago
5 0
84 cents divided by 14 inches = 6 cents per inch
48 cents divided by 6 cents = 8 inches

8 inches of wire can be bought for 48 cents
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Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

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Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

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Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

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