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Pavlova-9 [17]
2 years ago
10

On the first day of spring, an entire field of flowering trees blossoms. The population of locusts consuming these flowers rapid

ly increases as the trees blossom. The locust population increases by a factor of 5 every 22 days, and can be modeled by a function, L, which depends on the amount of time, t (in days).
Before the first day of spring, there were 7600 locusts in the population.


Write a function that models the locust population t days since the first day of spring.
Mathematics
1 answer:
Anon25 [30]2 years ago
5 0

Answer:

y=7600(5^(t/22))

Step-by-step explanation:

This is going to be an exponential function as it grows rapidly.

This type of question can be solved using the formula y=a(r^x), where a is the inital amount, r the factor by which the amount increases and x is the unit of time after which the amount increases.

x=t/22

a=7600

r=5

∴y=7600(5^(t/22))

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Step-by-step explanation:

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2 years ago
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An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t
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Answer:

35.2% probability that the sample mean will be 246 pages or more

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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 245 \sigma = 15, n = 33, s = \frac{15}{\sqrt{33}} = 2.61

What the probability that the sample mean will be 246 pages or more?

This is 1 subtracted by the pvalue of Z when X = 246. So

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

By the Central Limit Theorem

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

Z = \frac{246 - 245}{2.61}

Z = 0.38

Z = 0.38 has a pvalue of 0.6480.

1 - 0.6480 = 0.3520

35.2% probability that the sample mean will be 246 pages or more

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2 years ago
The equation r = 6t gives the number of albums released, r, by a record label over time in years, t. If you graph this relations
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Choices:
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(9, 52)
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r = 6t 
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(8,48) : r = 6(8) = 48 CORRECT
(9,52) : r = 6(9) = 54 incorrect
(12,74) : r = 6(12) = 72 incorrect
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The points that lie on the resulting line are (8,48) and (13,78)</span>
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The number of acres in a region cleared for farming follows the formula a = f (t) = 2t2 where t is the number of months since th
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we are given

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now, we can simplify it

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the average rate of change in the number of acres cleared for farming between t = 1 and t = 4  is 10 acres/ month.............Answer

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