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Natalka [10]
3 years ago
8

Which of the following numbers is a composite number? A. 11 B. 17 C. 23 D. 27

Mathematics
2 answers:
Advocard [28]3 years ago
8 0

Answer:

D 27

Step-by-step explanation:

A composite number is a number that is formed by multiplying two numbers together that are not itself and 1

A. 11

B. 17

C. 23

D. 27  We know that 27 = 3*9  so this number is composite

dimulka [17.4K]3 years ago
5 0

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                  Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

❖ A composite number is 27. 27 is composite because there are more numbers than 1 and itself that can be multiplied to create 27. The two other numbers to make 27 are 3 x 9. All the other answer choices are prime numbers because no other numbers besides 1 and itself can be multiplied to make that number.

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

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3 years ago
The amount of coffee that a filling machine puts into an 8-ounce jar is normally distributed with a mean of 8.2 ounces and a sta
nordsb [41]

Answer:

73.30% probability that the sampling error made in estimating the mean amount of coffee for all 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 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 = 8.2, \sigma = 0.18, n = 100, s = \frac{0.18}{\sqrt{100}} = 0.018

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Z = -1.11

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0.8665 - 0.1335 = 0.7330

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