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Grace [21]
3 years ago
5

.) Suppose college students produce 650 pounds of solid waste each year, on average. Assume that the distribution of waste per c

ollege student is normal with a mean of 650 pounds and a standard deviation of 20 pounds. What is the probability that a randomly selected student produces either less than 620 or more than 700 pounds of solid waste
Mathematics
1 answer:
amid [387]3 years ago
4 0

Answer:

0.073 = 7.3% probability that a randomly selected student produces either less than 620 or more than 700 pounds of solid waste

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 650 pounds and a standard deviation of 20 pounds.

This means that \mu = 650, \sigma = 20

What is the probability that a randomly selected student produces either less than 620 or more than 700 pounds of solid waste?

Less than 620:

pvalue of Z when X = 620. So

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

Z = \frac{620 - 650}{20}

Z = -1.5

Z = -1.5 has a pvalue of 0.0668

More than 700:

1 subtracted by the pvalue of Z when X = 700. So

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

Z = \frac{700 - 650}{20}

Z = 2.5

Z = 2.5 has a pvalue of 0.9938

1 - 0.9938 = 0.0062

Total:

0.0668 + 0.0062 = 0.073

0.073 = 7.3% probability that a randomly selected student produces either less than 620 or more than 700 pounds of solid waste

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Step-by-step explanation:

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A service station will be built on the highway, and a road will connect it with cray. How long will the new road be?
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Answer:

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Step-by-step explanation:

In this question we will follow the property of congruence of two triangles.

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As we know in the congruent triangles all sides are in the same ratio.

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\frac{Distance(Alba-Cray)}{Distance(Alba-Blare)}= \frac{Distance(Cray-Service station)}{Distance(Cray-Blare)}

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3 years ago
Complete the square and write in standard form. Show all work.What would be the conic section:CircleEllipseHyperbolaParabola
mote1985 [20]

ANSWER

This is an ellipse. The equation is:

\frac{(x-1)^2}{3^2}+\frac{(y+4)^2}{4^2}=1

EXPLANATION

We have to complete the square for each variable. To do so, we have to take the first two terms and compare them with the perfect binomial squared formula,

(a+b)^2=a^2+2ab+b^2

For x we have to take 16x² and -32x. Since the coefficient of x is not 1, first, we have to factor out the coefficient 16,

16x^2-32x=16(x^2-2x)

Now, the first term of the expanded binomial would be x and the second term -2x. Thus, the binomial is,

(x-1)^2=x^2-2x+1

To maintain the equation, we have to subtract 1,

16(x^2-2x+1-1)=16((x-1)^2-1)=16(x-1)^2-16

Now, we replace (16x² - 32x) from the given equation by this equivalent expression,

16(x-1)^2-16+9y^2+72y+16=0

The next step is to do the same for y. We have the terms 9y² + 72y. Again, since the coefficient of y² is not 1, we factor out the coefficient 9,

9y^2+72y=9(y^2+8y)

Following the same reasoning as before, we have that the perfect binomial squared is,

(y+4)^2=y^2+8y+16

Remember to subtract the independent term to maintain the equation,

9(y^2+8y)=9(y^2+8y+16-16)=9((y+4)^2-16)=9(y+4)^2-144

And now, as we did for x, replace the two terms (9y² + 72y) with this equivalent expression in the equation,

16(x-1)^2-16+9(y+4)^2-144+16=0

Add like terms,

\begin{gathered} 16(x-1)^2+9(y+4)^2+(-16-144+16)=0 \\ 16(x-1)^2+9(y+4)^2-144=0 \end{gathered}

Add 144 to both sides,

\begin{gathered} 16(x-1)^2+9(y+4)^2-144+144=0+144 \\ 16(x-1)^2+9(y+4)^2=144 \end{gathered}

As we can see, this is the equation of an ellipse. Its standard form is,

\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1

So the next step is to divide both sides by 144 and also write the coefficients as fractions in the denominator,

\begin{gathered} \frac{16(x-1)^2}{144}+\frac{9(y+4)^2}{144}=\frac{144}{144} \\  \\ \frac{(x-1)^2}{\frac{144}{16}}+\frac{(y+4)^2}{\frac{144}{9}}=1 \end{gathered}

Finally, we have to write the denominators as perfect squares, so we identify the values of a and b. 144 is 12², 16 is 4² and 9 is 3²,

\frac{(x-1)^2}{(\frac{12}{4})^2}+\frac{(y+4)^2}{(\frac{12}{3})^2}=1

Note that we can simplify a and b,

\frac{12}{4}=3\text{ and }\frac{12}{3}=4

Hence, the equation of the ellipse is,

\frac{(x-1)^2}{3^2}+\frac{(y+4)^2}{4^2}=1

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1 year ago
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