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1 year ago

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-5, 20, -80, 320, ... is a geometric sequence a. Find the common ratio, b. Write its recursive rule c. Write its explicit rule.W

hat is the common ratio and explicit rule?

1 answer:

mario62 [17]1 year ago

5 0

a.

In order to find the common ratio, we just need to divide a term by the term that comes before it.

So using the terms 20 and -5, we have:

$\text{ratio}=\frac{20}{-5}=-4$

b.

The recursive rule can be found with the formula:

$a_n=a_{n-1}\cdot q$

Where an is the nth term and q is the ratio. So we have:

$a_n=a_{n-1}\cdot(-4)$

c.

The explicit rule can be written as:

$a_n=a_1\cdot q^{n-1}$

Where an is the nth term, a1 is the first term and q is the ratio. So:

$a_n=-5\cdot(-4)^{n-1}$

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

$z = \frac{x-\mu }{\sigma }$

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

So

$P(x \ge 60) \approx P(z > 1.34)$

$= 1-P(z \le 1.34)$

$=1-0.9099$

$=0.0901$

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Step-by-step explanation:

From the question, we have the following information:

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We have here that protein requirements in adults is normally distributed with defined parameters. The question is about <em>the proportion</em> <em>of the population</em> that has a requirement less than $\\ x = 0.60 gP*kg^{-1}*d^{-1}$ .

For answering this, we need to calculate a <em>z-score</em> to obtain the probability of the value <em>x </em>in this distribution using a <em>standard normal table</em> available on the Internet or on any statistics book.

<h3>z-score</h3>

A z-score is expressed as

$\\ z = \frac{x - \mu}{\sigma}$

For the given parameters, we have:

$\\ z = \frac{0.60 - 0.65}{0.07}$

$\\ z = \frac{0.60 - 0.65}{0.07}$

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<h3>Determining the probability</h3>

With this value for <em>z</em> at hand, we need to consult a standard normal table to determine what the probability of this value is.

The value for z = -0.7142857 is telling us that the requirement for protein is below the population mean (negative sign indicates this). However, most standard normal tables give a probability that a statistic is less than z and for values greater than the mean (in other words, positive values). To overcome this, we need to take the complement of the probability given for z-score z = 0.7142857, that is, subtract from 1 this probability, which is possible because the normal distribution is <em>symmetrical</em>.

Tables have values for <em>z</em> with two decimal places, then, for z = 0.7142857, we need to rewrite it as z = 0.71. For this value, the <em>standard normal table</em> gives a value of P(z<0.71) = 0.76115.

Therefore, the cumulative probability for values less than x = 0.60 which corresponds to a z-score = -0.7142857 is approximately:

$\\ P(x$

$\\ P(x$ (rounding to three decimal places)

That is, the proportion of the population that have a protein requirement less than 0.60 g P • kg-1 • d-1 is

$\\ 0.239 =\frac{239}{1000}\;or\;23.9\%$

See the graph below. The shaded area is the region that represents the proportion asked in the question.

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