Explain the process for finding the product of two integers

2 answers:

8 0

The process of finding the product of two integers is multiplying them. A product is the answer to a multiplication problem.

E.g.:

Integers: 5 and 2

5*2 or 5 x 2
=10

Product: 10

Westkost [7]3 years ago

5 0

The <em>correct answer</em> is:

To multiply integers, you multiply the numbers as if they were whole numbers. When you are finished, you assign a sign to them using the following rules:

A negative multiplied by a negative is a positive;

A positive multiplied by a positive is a positive;

A positive multiplied by a negative is a negative;

A negative multiplied by a positive is a negative.

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Can someone please help quick!

Jet001 [13]

Answer:

the answer is 8

Step-by-step explanation:

1/2(8 + 4) = 6

0.5(8 + 4) = 6

(4 + 2) = 6

6 = 6

7 0

3 years ago

An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t

Neko [114]

Answer:

35.2% probability that the sample mean will be 246 pages or more

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 $s = \frac{\sigma}{\sqrt{n}}$ .