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Dmitrij [34]
3 years ago
14

If the number of bacteria in a colony doubles every 25 hours and there is currently a population of 10,000 bacteria, what will t

he population be 75 hours from now
Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
5 0

Answer:

The population of 75 hours from now would be 40,000.

Step-by-step explanation:

The reason for this is because your initial number is 10,000 and when 25 hours pass by, your population would be 20,000. Another 25 hours pass by, and then your population would be 40,000. If you multiply 25 by 3, you will notice 75 hours have just passed, so your answer would be 40,000.

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A new business has total sales of $10,000, applied $1,000 in merchandise discounts, and had returns of $100. Calculate the busin
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10,000-1,000=9,000 -100
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Can someone pls help anyone?
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40 units

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The mean life of a television set is 119 months with a standard deviation of 14 months. If a sample of 74 televisions is randoml
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Answer:

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.63

If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 months

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{120.1 - 119}{1.63}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

Z = \frac{X - \mu}{s}

Z = \frac{117.9 - 119}{1.63}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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