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galina1969 [7]
3 years ago
12

An archery target consists of a circular bull's-eye with radius x, surrounded by four rings with width y. What is the area of th

e outermost ring in terms of x and y?
Mathematics
1 answer:
amid [387]3 years ago
4 0
<span>given:
   bull's eye radius= x
 width of surrounding rings=y

   solution:
   Radius of the circle=x+4y
Area of the outermost ring=Area of the circle-Area of the penultimate ring =Ď€(x+4y)^2-Ď€(x+3y)^2
=Ď€(x^2+8xy+16y^2-x^2-9y^2-6xy)
 =Ď€(2xy+7y^2)
 hence the area of the outermost ring in terms of x and y is Ď€(2xy+7y^2).</span>
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Perform the indicated operation. Simplify the result in factored form. ((x - y)/ (x^2 - 1)) * ((x - 1)/(x^2 - y^2))
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Answer:

The answer to your question is            \frac{1}{(x + 1)(x + y)}            

Step-by-step explanation:

Real operation

                      \frac{x - y}{x^{2} - 1} \frac{x - 1}{x^{2} - y^{2}}

Process

1.- Factor the denominators

                     \frac{x - y}{(x + 1)(x - 1)} \frac{x - 1}{(x + y)(x - y)}

2.- Simplify like terms

Cancel (x - y) and (x - 1) because they are in numerator and the denominator

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3.- Expand

                   \frac{1}{x^{2} + xy + x + y^{2}}                    

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The weight of crackers in a box is stated to be 16 ounces. The amount that the packaging machine puts in the boxes is believed t
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Answer:

99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces

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 problem, we have that:

\mu = 16.15, \sigma = 0.3, n = 50, s = \frac{0.3}{\sqrt{50}} = 0.0424

What is the probability that the mean weight of a 50-box case of crackers is above 16 ounces?

This is 1 subtracted by the pvalue of Z when X = 16. So

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

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Z = \frac{16 - 16.15}{0.0424}

Z = -3.54

Z = -3.54 has a pvalue of 0.0001

1 - 0.0001 = 0.9999

99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces

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I would say C because it would make the most sense because it doesn't describe or state the other two choices given .
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4 years ago
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