A printer can print 12 pages in 9 seconds. What is the closest estimate of the

Mathematics

2 answers:

Volgvan3 years ago

8 0

80 in a minute

60 divided by 9
Times that answer by 12
Good luck!

ExtremeBDS [4]3 years ago

5 0

Answer:

Step-by-step explanation:

there are 60 seconds in a minute so about 6 9 seconds in a minute. then there is 12 pages per 9 seconds and since there is 6 9 seconds then you can just multiply 6 by 9 to get the answer 72

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The measure of one angle of a right triangle is 56∘ more than the measure of the smallest angle. Find the measure of the smalles

GrogVix [38]

Answer:

The sum of all angels for any triangle equals 180°

so we have a right triangle which means we have an angle equals 90° ,angle equals=56°.

step by step

$180 - 56 - 90 = 34$

so the angle tends to equal 34°.

7 0

3 years ago

Based on historical data, your manager believes that 37% of the company's orders come from first-time customers. A random sample

fomenos

Answer:

0.6214 = 62.14% probability that the sample proportion is between 0.26 and 0.38

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 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.

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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$