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boyakko [2]
3 years ago
5

In a certain​ country, the true probability of a baby being a boy is 0.524. Among the next five randomly selected births in the​

country, what is the probability that at least one of them is a girl​?
Mathematics
1 answer:
KonstantinChe [14]3 years ago
7 0

Answer:

The probability that at least one of them is a girl is 0.96049.

Step-by-step explanation:

In a certain​ country, the true probability of a baby being a boy is 0.524.

This is a binomial problem.

p(boy) = 0.524

n = 5

p(girl) = 1-0.524=0.476

P(at least one girl) = 1 - p(5 boys) =1-0.524^5 = 0.96049

Hence, the probability is 0.96049.

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Carly buys a gold ring priced at $273.00 if the sale tax is 10 % , how much will carly pay
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Hello!

<h2>Answer:</h2>

Carla will pay $300.30 for the gold ring.

<h2>Explanation:</h2>

Sales tax is added on to the total cost of the item. We need to figure out how much tax is on the ring.

To do that, we must calculate 10% (or 0.10) of $273.00.

273 × 0.10 = 27.3

Now, add the tax to the cost of the ring.

273 + 27.3 = 300.3

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2 years ago
What is the simplified value of the expression below?-8 *(-3)
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Does anybody know how to do time with exponential decay
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<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

and so the amount of air remaining after 50 breaths is

b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}

which is a very small number close to zero.</span>
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