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Anastasy [175]
1 year ago
15

We have a shape that has an area of 112 inches squared. Convert the units to centimeters squared. (HINT: Perform multiple conver

sions). Don’t forget to show your work.
Mathematics
1 answer:
Andre45 [30]1 year ago
5 0
<h3>Answer: 722.5792 square centimeters</h3>

Work Shown:

112 \text{ in}^2 = 112 \text{ in}^2 \times \frac{2.54 \text{ cm}}{1 \text{ in}} \times \frac{2.54 \text{ cm}}{1 \text{ in}}\\\\112 \text{ in}^2 = (112 \times 2.54 \times 2.54) \text{ cm}^2\\\\112 \text{ in}^2 = 112 \times (2.54)^2 \text{ cm}^2\\\\112 \text{ in}^2 = 722.5792 \text{ cm}^2\\\\

Notes:

  • 1 inch = 2.54 cm exactly
  • The conversion factor \frac{2.54 \text{ cm}}{1 \text{ in}} is used to convert from inches to cm. The inches unit in the denominator cancels with one of the inches unit from the original 112 square inches figure.
  • We use two identical copies of the same conversion factor to convert from square inches to square centimeters. This is the "multple conversion" your teacher is referring to.
  • Think of "square inches", aka \text{in}^2, as "inches times inches". Or think of a 1 inch by 1 inch square that has an area of 1 square inch. The same applies for the square centimeter unit as well.
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Suppose that the population mean for income is $50,000, while the population standard deviation is 25,000. If we select a random
Fudgin [204]

Answer:

Probability that the sample will have a mean that is greater than $52,000 is 0.0057.

Step-by-step explanation:

We are given that the population mean for income is $50,000, while the population standard deviation is 25,000.

We select a random sample of 1,000 people.

<em>Let </em>\bar X<em> = sample mean</em>

The z-score probability distribution for sample mean is given by;

               Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean = $50,000

            \sigma = population standard deviation = $25,000

            n = sample of people = 1,000

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

So, probability that the sample will have a mean that is greater than $52,000 is given by = P(\bar X > $52,000)

  P(\bar X > $52,000) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{52,000-50,000}{\frac{25,000}{\sqrt{1,000} } } ) = P(Z > 2.53) = 1 - P(Z \leq 2.53)

                                                                    = 1 - 0.9943 = 0.0057

<em>Now, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.</em>

Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.

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