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Pachacha [2.7K]
3 years ago
6

What is the approximate circumference of Circle C? Use pi star times text   end text ≈ 3.14. Record your answer to the nearest h

undredth. m circle with a center point labeled C and a radius of four meters
Mathematics
1 answer:
Stels [109]3 years ago
7 0
Circumference of a circle is found with the equation

C = 2 × pi × r

where r is the radius. So for your problem...

C = 2 × 3.14 × 4
C = 25.12 meters
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PLEASE HELP ASAP!! THANKS SO MUCH
Wittaler [7]

Answer:

The relationship shows a direct linear variation with a constant of variation of 1.

Step-by-step explanation:

This is true because the slope of the equation is 1

4 0
3 years ago
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Draw two lines with a slope of 1 / 2. What do you notice about the lines?
patriot [66]

Answer:

The lines would be parallel.

Step-by-step explanation:

lines with the same slope are parallel.

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3 years ago
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Using the weighted average approach to process costing, floridyne would use what number of equivalent units in 2019 to calculate
timofeeve [1]

Floridyne would use an equivalent unit labor of 166,340 unit to calculate the cost per equivalent unit for direct labor.

<h3>How do we get the equivalent unit labor?</h3>

Under weighted average, we do not make distinction between started and finished and just finished. Thus we work with finished and ending WIP only:

Finished         162,000

Ending - WIP    6,200 ending at 70% complete

Equivalent units for labor = Finished + Percentage of completion ending units

= 162,000 + 6,200 x 70%

= 166,340

Therefore, he would use an equivalent unit labor of 166,340 unit to calculate the cost per equivalent unit for direct labor.

Missing word "Floridyne, Inc. manufactures mouthwash. They had no finished goods inventory at the beginning of 2019. They have only one processing department for this product. A review of the company’s inventory records shows the following: At the beginning of January 2019, Floridyne has 4,500 gallons of mouthwash in process. (costs $8,410 for materials, 1,663 for labor and 4,990 for overhead) During 2019, Floridyne finishes/transfers 162,000 gallons of mouthwash. On December 31, 2019, Floridyne has 6,200 gallons of mouthwash that is 70% complete. Direct materials are added half at the beginning of the process and half after the process is 60% complete. During 2019 $349,000 of direct materials and $92,500 of direct labor were added. Using the weighted average approach to process costing,"

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4 0
2 years ago
Find the mean, median, mode, and range of the data set.
VikaD [51]
Mode: 23

Range: 9

Median: 25.5

Mean: 25.5625
5 0
3 years ago
Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1
svet-max [94.6K]

Answer:

a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 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 normal distributions can be 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.

Mean of 8.0 ounces and a standard deviation of 1.1 ounces.

This means that \mu = 8, \sigma = 1.1

(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?

n = 5 means that s = \frac{1.1}{\sqrt{5}} = 0.4919

This probability is the pvalue of Z when X = 9.3. So

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

By the Central Limit Theorem

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

Z = \frac{9.3 - 8}{0.4919}

Z = 2.64

Z = 2.64 has a pvalue of 0.9959

0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

(b) If 6 potatoes are randomly selected, find the probability that the mean weight is more than 9.0 ounces?

n = 6 means that s = \frac{1.1}{\sqrt{6}} = 0.4491

This probability is 1 subtracted by the pvalue of Z when X = 9. So

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

Z = \frac{9 - 8}{0.4491}

Z = 2.23

Z = 2.23 has a pvalue of 0.9871

1 - 0.9871 = 0.0129

0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

8 0
3 years ago
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