What is a mixed number greater than one

Order the data from lest to greatest 35, 24,25,38,31,20,27

kolbaska11 [484]

Answer:

Step-by-step explanation:

20,24,25,27,31,35,38

minimum;20

medium: 27

maximum: 38

lower quartile: 24

upper quartile:35

put numbers in number order

the first number is the minimum, the last number is the maximum. the middle number "split" is the median.

quartiles are the numbers in the middle of the data you just "split"

(so like a quarter)

i may be wrong i haven't done this in years

3 0

2 years ago

08.01) Which of the following statements shows a characteristic of a statistical question? (4 points) Select one: a. The questio

Alexeev081 [22]

Answer:

the distribution can be broken into categories

7 0

2 years ago

Read 2 more answers

Solve/answer the question and help me understand this question please

faltersainse [42]

Answer:

5.25 m

Step-by-step explanation:

A diagram can help you understand the question, and can give you a clue as to how to find the answer. A diagram is attached. The problem can be described as finding the sum of two vectors whose magnitude and direction are known.

__

<h3>understanding the direction</h3>

In navigation problems, direction angles are specified a couple of different ways. A <em>bearing</em> is usually an angle in the range [0°, 360°), <em>measured clockwise from north</em>. In land surveying and some other applications, a bearing may be specified as an angle east or west of a north-south line. In this problem we are given the bearing of the second leg of the walk as ...

N 35° E . . . . . . . 35° east of north

Occasionally, a non-standard bearing will be given in terms of an angle north or south of an east-west line. The same bearing could be specified as E 55° N, for example.

<h3>the two vectors</h3>

A vector is a mathematical object that has both magnitude and direction. It is sometimes expressed as an ordered pair: (magnitude; direction angle). It can also be expressed using some other notations;

magnitude∠direction
magnitude <em>cis</em> direction

In the latter case, "cis" is an abbreviation for the sum cos(θ)+i·sin(θ), where θ is the direction angle.

Sometimes a semicolon is used in the polar coordinate ordered pair to distinguish the coordinates from (x, y) rectangular coordinates.

__

The first leg of the walk is 3 meters due north. The angle from north is 0°, and the magnitude of the distance is 3 meters. We can express this vector in any of the ways described above. One convenient way is 3∠0°.

The second leg of the walk is 2.5 meters on a bearing 35° clockwise from north. This leg can be described by the vector 2.5∠35°.

<h3>vector sum</h3>

The final position is the sum of these two changes in position:

3∠0° +2.5∠35°

Some calculators can compute this sum directly. The result from one such calculator is shown in the second attachment:

= 5.24760∠15.8582°

This tells you the magnitude of the distance from the original position is about 5.25 meters. (This value is also shown in the first attachment.)

__

You may have noticed that adding two vectors often results in a triangle. The magnitude of the vector sum can also be found using the Law of Cosines to solve the triangle. For the triangle shown in the first attachment, the Law of Cosines formula can be written as ...

a² = b² +o² -2bo·cos(A) . . . . where A is the internal angle at A, 145°

Using the values we know, this becomes ...

a² = 3² +2.5² -2(3)(2.5)cos(145°) ≈ 27.5373

a = √27.5373 = 5.24760 . . . . meters

The distance from the original position is about 5.25 meters.

_____

<em>Additional comment</em>

The vector sum can also be calculated in terms of rectangular coordinates. Position A has rectangular coordinates (0, 3). The change in coordinates from A to B can be represented as 2.5(sin(35°), cos(35°)) ≈ (1.434, 2.048). Then the coordinates of B are ...

(0, 3) +(1.434, 2.048) = (1.434, 5.048)

The distance can be found using the Pythagorean theorem:

OB = √(1.434² +5.048²) ≈ 5.248

7 0

1 year ago

(a) If G is a finite group of even order, show that there must be an element a = e, such that a−1 = a (b) Give an example to sho

Dahasolnce [82]

Answer:

See proof below

Step-by-step explanation:

First, notice that if a≠e and a^-1=a, then a²=e (this is an equivalent way of formulating the problem).

a) Since G has even order, |G|=2n for some positive number n. Let e be the identity element of G. Then A=G\{e} is a set with 2n-1 elements.

Now reason inductively with A by "pairing elements with its inverses":

List A as A={a1,a2,a3,...,a_(2n-1)}. If a1²=e, then we have proved the theorem.

If not, then a1^(-1)≠a1, hence a1^(-1)=aj for some j>1 (it is impossible that a^(-1)=e, since e is the only element in G such that e^(-1)=e). Reorder the elements of A in such a way that a2=a^(-1), therefore a2^(-1)=a1.

Now consider the set A\{a1,a2}={a3,a4,...,a_(2n-1)}. If a3²=e, then we have proved the theorem.

If not, then a3^(-1)≠a1, hence we can reorder this set to get a3^(-1)=a4 (it is impossible that a^(-1)∈{e,a1,a2} because inverses are unique and e^(-1)=e, a1^(-1)=a2, a2^(-1)=a1 and a3∉{e,a1,a2}.

Again, consider A\{a1,a2,a3,a4}={a5,a6,...,a_(2n-1)} and repeat this reasoning. In the k-th step, either we proved the theorem, or obtained that a_(2k-1)^(-1)=a_(2k)

After n-1 steps, if the theorem has not been proven, we end up with the set A\{a1,a2,a3,a4,...,a_(2n-3), a_(2n-2)}={a_(2n-1)}. By process of elimination, we must have that a_(2n-1)^(-1)=a_(2n-1), since this last element was not chosen from any of the previous inverses. Additionally, a_(2n1)≠e by construction. Hence, in any case, the statement holds true.

b) Consider the group (Z3,+), the integers modulo 3 with addition modulo 3. (Z3={0,1,2}). Z3 has odd order, namely |Z3|=3.

Here, e=0. Note that 1²=1+1=2≠e, and 2²=2+2=4mod3=1≠e. Therefore the conclusion of part a) does not hold

7 0

3 years ago

Plz helpp

den301095 [7]

Answer:

Addition Property of Equality If AB = CD then AB + BC = BC +CD

Subtraction Property of Equality If AB + BC = BC + CD then AB = CD

Multiplication Property of Equality If m∢A = 90 then 2(m∢A) = 180

Division Property of Equaliity If 2(m∢B) = 180 then m∢B = 90

Substitution Property If m∢A + m∢B =180 and m∢B then m∢A + 90 =180

Distributive Property AB + AB = 2AB

Reflexive Property m∢B = m∢B

Symmetric Property If AB + BC = AC then AC = AB + BC

Transitive Property If AB ≅ BC and BC ≅ CD then AB ≅ CD

Segment Addition Postulate If C is between B and D, then BC + CD = BD

Angle Addition Postulate If D is a point in the interior of ∢ABC then m∢ABD + m∢DBC = m∢ABC

Linear Pair Postulate If two angles form a linear pair, then they are supplementary

Definition of Right Angle If ∢B is a right angle then m∢B = 90

Definition of Midpoint If P is the midpoint of segment AB then AP =PB

Definition of Segment Bisector If k intersects segment AB at M the Midpoint then k bisects segment AB

Definition of Perpendicular Lines If two lines are ⊥ they form right angles

Definition of Congruent Segments If AB = CD then segment AB ≅ segment CD

Definition of Congruent Angles If ∡A ≅∡ B then m∡A=m∡B

Definition of Angle Bisector If ray AB bisects ∡CAD then∡ CAB ≅ ∡ BAD

Definition of Complementary Angles If ∡ Z and ∡Y are complementary m∡Z +m∡Y =90

Definition of Supplementary Angles If ∡ S and ∡T are supplementary m∡S +m∡T = 180

Step-by-step explanation:

5 0

3 years ago