Is 6.5892 a rational or irrational number

Answers

Answer 1
Answer:

The number, 6.5892 , given that it ends, which fits the definition, is therefore a rational number.

What are rational numbers ?

A rational number is a number that can be expressed as a fraction of two integers, where the denominator is not 0. 6.5892 can be expressed as the fraction 65892/10000, which is a simplified fraction with an integer numerator and denominator.

An irrational number is a number that cannot be expressed as a fraction of two integers. Examples of irrational numbers include pi (π) and the square root of 2.

A simple reminder, Rational numbers can be written as fractions. Irrational numbers cannot be written as fractions.

Find out more on rational numbers at brainly.com/question/19079438

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Answer 2
Answer: 6.5892 is a irrational number

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Simplify \sqrt{25a^2}

Are the rational numbers closed under multiplication?

Answers

Answer:

Yes, The rational numbers are closed under multiplication.

Step-by-step explanation:

A rational number is a number which can be expressed in the form of a fraction (x)/(y), where x and y are integers and y ≠ 0.

Now, closure property of multiplication states that if two rational numbers are multiplied then the product is also a rational number. Thus, if r and t are rational numbers, then

r×t = s, where s is the product of r and t

s is also a rational number.

Hence, the rational numbers are closed under multiplication.

This can be better explained with the help of an example (3)/(4) * (2)/(5) = (6)/(20),

It is clear that (6)/(20) is a rational number.

How do you find the scale factor

Answers

Scale factor(smaller length) is larger length over smaller length. Scale factor(larger length) is smaller length over larger length. Add your measurement and your done!

Q # 18,Graph the inequality on a coordinate plane, - y < 3 x - 5

Answers

For this case we have the following inequality:
 - y <3 x - 5
 Rewriting we have:
 y > -3x + 5
 The solution is given in this case by the set of points that belong to the shaded region shown in the graph.
 Answer:
 
see attached image.

An educational psychologist wants to test whether a new teaching method negatively affects reading comprehension scores. She randomly selects 30 6th grade students that were taught under the new teaching method and finds that their scores on a standardized reading comprehension test have a mean equal to 118.8 with a variance equal to 37.2. Scores on the standardized test in the general population of 6th graders are distributed approximately normally with a mean equal to 119.8. Is there sufficient evidence to conclude that the new teaching method negatively affects reading comprehension scores

Answers

Answer:

We conclude that the new teaching method does not negatively affects reading comprehension scores.

Step-by-step explanation:

We are given that an educational psychologist wants to test whether a new teaching method negatively affects reading comprehension scores.

She randomly selects 30 6th grade students that were taught under the new teaching method and finds that their scores on a standardized reading comprehension test have a mean equal to 118.8 with a variance equal to 37.2.

Scores on the standardized test in the general population of 6th graders are distributed approximately normally with a mean equal to 119.8.

Let \mu = mean scores on a standardized reading comprehension test.

So, Null Hypothesis, H_0 : \mu \geq 119.8      {means that the new teaching method does not negatively affects reading comprehension scores}

Alternate Hypothesis,H_A : \mu < 119.8    {means that the new teaching method negatively affects reading comprehension scores}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

                      T.S. =  (\bar X-\mu)/((s)/(√(n) ) )  ~ t_n_-_1

where, \bar X = sample mean test score = 118.8

            s = sample standard deviation = √(37.2) = 6.1

            n = sample of 6th grade students = 30

So, test statistics  =  (118.8-119.8)/((6.1)/(√(30) ) )  ~ t_2_9

                              =  -0.898

The value of t test statistics is -0.898.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the t table gives critical value of -1.699 for left-tailed test.

Since our test statistic is more than the critical value of t as -0.898 > -1.699, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the new teaching method does not negatively affects reading comprehension scores.

Are the points B, F, and W coplanar?​

Answers

Answer:

yes

Step-by-step explanation:

Because the point lie on the plane they are coplanar

How many randomly selected employers must we contact in order to create an estimate in which we are 95​% confident with a margin of error of 9​%? ​b) Suppose we want to reduce the margin of error to 4​%. What sample size will​ suffice? ​c) Why might it not be worth the effort to try to get an interval with a margin of error of 1​%?

Answers

Answer:

a)n=543

b)n=1509

c)n=13573

Step-by-step explanation:

a)

c=98%,

E=0.05

Margin Error E=Zα/2√p(1-p)/n

but n=((Zα/2)/n)²×p(1-p)

where the confidence level is 1-α=0.98

cross multiply

Zα/2=2.33

where p=0.5

input the values

n=(2.33/0.55)²×0.5(1-0.5)=543

n=0.33

b) E=0.33

E=Zα/2√p(1-p)/n

n=((Zα/2)/n)²×p(1-p)

1-α=0.01 confidence level

n=(2.33/0.33)²×0.5(1-0.5)=1504

n=1504

c) E=Zα/2√p(1-p)/n

n=((Zα/2)/n)²×p(1-p)

1-α=0.98

cross multiply

Zα/2=2.33

p=0.5

n=(2.33/0.01)²×0.5(1-0.5)=13573

n=13573