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2 years ago

List all two-digit positive integers that satisfy both of the following :

2 answers:

6 0

The two-digit positive integers that tens and one digits are consecutive numbers include 12, 23, 34, 45, 56, 67, 78, and 89.

We have given that,

1. The tens and one's digits are consecutive numbers, and

2. The integer is the product of two consecutive numbers.

<h3>What are integers?</h3>

It should be noted that integers are the whole numbers that can be either positive, negative, or zero.

The two-digit positive integers that the integer is the product of two consecutive numbers will be:

3 × 4 = 12

7 × 8 = 56

Learn more about integers on:

brainly.com/question/17695139

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MakcuM [25]2 years ago

3 0

The two-digit positive integers that tens and ones digits are consecutive numbers include 12, 23, 34, 45, 56, 67, 78, and 89.

<h3>What are integers?</h3>

It should be noted that integers are the whole numbers that can be either positive, negative, or zero.

The two-digit positive integers that the integer is the product of two consecutive numbers will be:

3 × 4 = 12

7 × 8 = 56

Learn more about integers on:

brainly.com/question/17695139

#SPJ1

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Which of the following must be true about a perpendicular bisector and the segment it bisects? A. The perpendicular bisector and

Law Incorporation [45]

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4 years ago

Read 2 more answers

Simplify: −7 + {12 − 3[50 − 4(2 + 3) ] }

DedPeter [7]

Answer:

-85

Step-by-step explanation:

<em><u>1) (2 + 3)</u></em>

-7 + {12 - 3[50 - 4(5) ] }

<em><u>2) -7 + {12 - 3[50 - 20] }</u></em>

-7 + {12 - 3[30] }

<em><u>3) -7 + 12 - 90</u></em>

5 - 90

<u>Answer: -85</u>

6 0

3 years ago

Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame

lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

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:

$\mu = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678$

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

$Z = \frac{207.9 - 208}{0.1678}$