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2 years ago

Which represents a measure of volume? O 5 cm O 5 square cm 05 cm 05 cm

Mathematics

2 answers:

Elena-2011 [213]2 years ago

7 0

Answer:

a) 5 $cm^{3}$

Step-by-step explanation:

Bodies have three dimensions (width, height and depth). Measuring volume is calculating the number of cubic units that can fit inside.

When raising to 3, these dimensions are included and therefore 5 $cm^{3}$ is a measure of volume.

tester [92]2 years ago

7 0

Answer:

5cm^3

Step-by-step explanation:

Volume for a form will always be in cubic units.

Area of a shape will always be in squared units.

Length will not be in cubic or squared units.

Hence, the first option is a measure for volume.

The second option is equal to 5cm^2 which represents a measure for area, as does the fourth option.

The third option represents a measurement of length, for example the length of a line segment or the height of a figure.

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2 years ago

Read 2 more answers

The amount of coffee that a filling machine puts into an 8 dash ounce 8-ounce jar is normally distributed with a mean of 8.2 oun

Inessa [10]

Answer:

73.3% probability that the sampling error made in estimating the mean amount of coffee for all 8 dash ounce 8-ounce jars by the mean of a random sample of 100 jars, will be at most 0.02 ounce

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theore.

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 question, we have that:

$\mu = 8.2, \sigma = 0.18, n = 100, s = \frac{0.18}{\sqrt{100}} = 0.018$

What is the probability that the sampling error made in estimating the mean amount of coffee for all 8 dash ounce 8-ounce jars by the mean of a random sample of 100 jars, will be at most 0.02 ounce

That is, probability of the sample mean between 8.2-0.02 = 8.18 and 8.2 + 0.02 = 8.22, which is the pvalue of Z when X = 8.22 subtracted by the pvalue of Z when X = 8.18.

X = 8.22

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

By the Central Limit Theorem

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

$Z = \frac{8.22 - 8.2}{0.018}$

$Z = 1.11$

$Z = 1.11$ has a pvalue of 0.8665.

X = 8.18

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

$Z = \frac{8.18 - 8.2}{0.018}$

$Z = -1.11$

$Z = -1.11$ has a pvalue of 0.1335.

0.8665 - 0.1335 = 0.7330

73.3% probability that the sampling error made in estimating the mean amount of coffee for all 8 dash ounce 8-ounce jars by the mean of a random sample of 100 jars, will be at most 0.02 ounce

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3 years ago