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devlian [24]
3 years ago
7

Ricky is at the bakery shop when he sees that 8 pastries cost $160.he need 16 pastries.how much will 16 pastries cost?

Mathematics
2 answers:
IgorC [24]3 years ago
8 0
You can multiply $160 by 2 because 16 is 8 * 2. 16 pastries would then be $320.

Or, you can divide $160 by 8, to get the price per pastry. You would get $20 per pastry, and multiply it by 16, which equals $320

(You would get the same answer either way)
murzikaleks [220]3 years ago
4 0
16 pastries would cost the double cost of 8 pastries because 8 is half of 16
, if 8 pastries cost $160 then 16 pastries would cost $320.  
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Answer:

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b) 0.4219 = 42.19% probability that only one child will develop the disease.

c) 0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

Step-by-step explanation:

For each children, there are only two possible outcomes. Either they carry the disease, or they do not. The probability of a children carrying the disease is independent of any other children, which means that the binomial probability distribution is used 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)!}

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The probability that their offspring will develop the disease is approximately .25.

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Question a:

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P(X = 3) = C_{3,3}.(0.25)^{3}.(0.75)^{0} = 0.0156

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Question b:

This is P(X = 1). So

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

P(X = 1) = C_{3,1}.(0.25)^{1}.(0.75)^{2} = 0.4219

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c. The third child will develop Tay–Sachs disease, given that the first two did not.

Third independent of the first two, so just multiply the probabilities.

First two do not develop, each with 0.75 probability.

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p = 0.75*0.75*0.25 = 0.1406

0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

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