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beks73 [17]
3 years ago
15

the number of mosquito at the beginning of the summer was 4,000. the population of mosquito is expected to grow at a rate of 25%

a month . how many mosquitoes will there be after 4 months
Mathematics
1 answer:
lara31 [8.8K]3 years ago
3 0

Answer:

<em>about 9765 mosquitoes</em>

<em />

Step-by-step explanation:

This is an exponential growth problem which follows the formula:

A=P(1+r)^t

Where A is the future amount (we need this after 4 months)

P is the initial amount (here, it is 4000)

r is the rate of increase (here it is 25% or 0.25)

t is the time period (here t= 4, because we want to know after 4 months)

<em>plugging in all the available/given info, we can solve for A:</em>

<em>A=P(1+r)^t\\A=4000(1+0.25)^4\\A=4000(1.25)^4\\A=9765.63</em>

<em />

<em>Since, no fractional mosquitoes, we round down to </em><em>9765 mosquitoes</em>

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4 years ago
Find the slope of a line perpendicular to 2x-y-16
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3 years ago
A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40
bezimeni [28]

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

a. How large was the sample used in this survey?

We have that s = 25, \sigma = 400. We want to find n, so:

s = \frac{\sigma}{\sqrt{n}}

25 = \frac{400}{\sqrt{n}}

25\sqrt{n} = 400

\sqrt{n} = \frac{400}{25}

\sqrt{n} = 16

(\sqrt{n})^2 = 16^2[tex][tex]n = 256

A sample of 256 was used in this survey.

b. What is the probability that the point estimate was within ±15 of the population mean?

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Z = -15/25 = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

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