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Ksivusya [100]
2 years ago
9

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

Mathematics
2 answers:
Ghella [55]2 years ago
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

#SPJ1

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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Answer:

The 95​% confidence interval for the mean birth weight of all ​non-premature babies is between 3350.4 grams and 3662.4 grams. This means that we are 95% sure that true mean birth weight of all non-premature babies is in this interval.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

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Now, we have to find a value of T, which is found looking at the t table, with 40 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 2.0211

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.0211\frac{494.2}{\sqrt{41}} = 156

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 3506.4 - 156 = 3350.4 grams

The upper end of the interval is the sample mean added to M. So it is 3506.4 + 156 = 3662.4 grams

The 95​% confidence interval for the mean birth weight of all ​non-premature babies is between 3350.4 grams and 3662.4 grams. This means that we are 95% sure that true mean birth weight of all non-premature babies is in this interval.

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