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viva [34]
3 years ago
8

The Collatz conjecture is one of the most famous unsolved mathematical problems, because it's so simple, you can explain it to a

primary-school-aged kid, and they'll probably be intrigued enough to try and find the answer for themselves.
Mathematics
1 answer:
Vinil7 [7]3 years ago
6 0

Hi, you've asked an incomplete question. However, I inferred you need a brief explanation about The Collatz conjecture.

<u>Explanation:</u>

Put simply, what the Collatz conjecture unsolved problem entails is that if any positive number is picked and it is:

  1. An even number (eg 2, 4, 6,...), then if they are divided by 2,  the new number gotten should undergo the same process (that is to be divided by 2), it is believed your calculation would finally end up at 1. For example, let's pick the number 6, (6÷2=4; repeating the process 4÷2=<u>1</u>)
  2. An odd number, then if they are multiplied by 3 and 1 is added to the result, it is believed that your calculation would finally end up at 1.
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What is the standard form of y –7 = – 23 (x + 1)
ValentinkaMS [17]

Answer:

23x+y=-16

Step-by-step explanation:

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3 years ago
In rectangle ANHG, whose perimeter is 100, OP, PQ, and QR are congruent and mutually perpendicular and O is the midpoint of AN.
MariettaO [177]

Answer:

5

Step-by-step explanation:

The sum of adjacent sides of the rectangle is half the perimeter, 50, so ...

... AH = 50-40 = 10

Then ...

... OP +QR = 10 = 2×OP . . . . . QR ≅ OP

... OP = 5 = PQ . . . . . . . . . . . . PQ ≅ OP

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3 years ago
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from two points one on each leg of an isosceles triangle perpendicular are drawn to the base prove that the triangles formed are
puteri [66]

The description below proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.

<h3>How to prove an Isosceles Triangle?</h3>

Let ABC be an isosceles triangle such that AB = AC.

Let AD be the bisector of ∠A.

We want to prove that BD=DC

In △ABD & △ACD

AB = AC(Thus, △ABC is an isosceles triangle)

∠BAD =∠CAD(Because AD is the bisector of ∠A)

AD = AD(Common sides)

By SAS Congruency, we have;

△ABD ≅ △ACD

By corresponding parts of congruent triangles, we can say that; BD=DC

Thus, this proves that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.

Read more about Isosceles Triangle at; brainly.com/question/1475130

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4 0
2 years ago
A professor wants to know whether or not there is a difference in comprehension of a lab assignment among students depending on
san4es73 [151]

Answer:

It is concluded that no difference exists in the comprehension of the lab based on the test scores.

Step-by-step explanation:

<em><u>Let the populations be normally distributed with a populations standard deviation of 5.32 points for both the text and visual illustrations.</u></em>

The above statement tells us that the independent t or two sample t test will be performed as both have equal variances and are normally distributed.

This can be easily done through excel.

The following table is obtained

     

t-Test: Two-Sample Assuming Equal Variances  

 

                <u>  Text                      Visual Illustrations</u>

Mean          70.28                       75.08

Variance 304.48                     228.58

Observations 15                             15

Pooled Variance Sp²= 266.53

       Pooled Standard Deviation = Sp = 16.33

Hypothesized Mean Difference = x1`-x2`= 0

df = n1+n2-2= 15+15-2= 30-2= 28

t Stat -0.805188239

<u>P(T<=t) two-tail 0.427495979                      </u>

<u>t Critical two-tail 1.701130908                              </u>

Let the null and alternate hypotheses be

H0 : u1-u2= 0   against the claim Ha: u1-u2≠0

There is no difference between the means

against the claim

that there is a difference in means of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations.

The significance level is ∝= 0.1

The d.f is n1+n2-2= 15+15-2= 30-2=28

This is a two tailed test and the critical region is t (0.025) (28) ≥ 1.7011 and  t (0.025) (28) ≤ - 1.7011.

The test statistic is

t= x1-x2/ Sp √1/n1+ 1/n2

t= 70.28 -75.08/ 16.33√1/15 +1/15

t= -4.8/5.963

t= -0.8049811  ( minute difference from excel result due to rounding)

Since the calculated value of t= -0.8049 does not fall in the critical region we conclude that there is no difference in means of a lab assignment among students depending on if the instructions are given all in text, or if they are given primarily with visual illustrations.

We accept the null hypothesis.

The p- value is 0.427495979.

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1,229

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