1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
podryga [215]
3 years ago
13

Greek city-states began to form in 800 B.C. and in 200 A.D, Christians began getting persecuted. How many years passed between t

he time that Greek city-states began to form and the persecution of Christians began?
Mathematics
1 answer:
asambeis [7]3 years ago
6 0

Answer: 1000 years

Step-by-step explanation:

-Greek city-states began to form in 800 B.C i.e 800 years Before Christ appearance

- up till 200 A.D i.e 200 years of After Christ disappearance

Hence, the number of years between the time that Greek city-states began to form and the persecution of Christians began is equal to the sum of years Greek city States started formation and Year persecution began

i.e 800 BC + 200 A.D = 1000 Years

Thus, the number of years of that passed was 1000 years

You might be interested in
Construct the graph of the direct proportion y = kx for k = -1
Sergio [31]

Answer:

In le picture

8 0
3 years ago
A 2-column table with 5 rows. Column 1 is labeled Evelyn's Scores with entries 125, 137, 138, 145, 145. Column 2 is labeled Dist
padilas [110]

Answer:

Column 1(X)      Column 2(mean - X)

  125                   -13

  137                     -1

  138                     0

  145                     7

  145                     7

Mean = 138

Mean Absolute Deviation : 5.6

3 0
3 years ago
Read 2 more answers
How many soloutions does y=3x+1
katrin2010 [14]

Answer:

infinite points along the line

Step-by-step explanation:

This is the equation for a line.  A line has infinite points.  So there are infinite solutions along the line

3 0
3 years ago
Answer this please it’s for algebra 2
Sergeu [11.5K]

Part a)

P(junior) = (number of juniors)/(number total)

P(junior) = 235/705

P(junior) = 1/3 exactly

P(junior) = 0.33333 approximately

======================================

Part b)

P(freshman and LG) = (number of freshman who have LG)/(number total)

P(freshman and LG) = 70/705

P(freshman and LG) = 14/141 exactly

P(freshman and LG) = 0.09929 approximately

======================================

Part c)

n(junior) = number of juniors = 235

n(samsung) = number of people who have samsung = 274

n(junior and samsung) = number of juniors who have samsung = 93

----

n(junior or samsung) = n(junior)+n(samsung)-n(junior and samsung)

n(junior or samsung) = 235+274-93

n(junior or samsung) = 416

----

P(junior or samsung) = n(junior or samsung)/n(total)

P(junior or samsung) = 416/705 exactly

P(junior or samsung) = 0.59007 approximately

======================================

Part d)

n(sophomore and apple) = number of sophomores who have apple

n(sophomore and apple) = 80

n(apple) = number of students who have apple

n(apple) = 261

P(sophomore | apple) = n(sophomore and apple)/n(apple)

P(sophomore | apple) = 80/261 exactly

P(sophomore | apple) = 0.30651 approximately

======================================

Part e)

define the events

A = junior who has apple

B = junior who has samsung

C = person is a junior

n(A or B) = number of juniors who have apple or samsung

n(A or B) = n(A) + n(B) ... A and B assumed mutually exclusive

n(A or B) = 87+93

n(A or B) = 180

n(C) = 235

P( (Apple or Samsung) | Junior ) = n(A or B)/n(C)

P( (Apple or Samsung) | Junior ) = 180/235

P( (Apple or Samsung) | Junior ) = 36/47 exactly

P( (Apple or Samsung) | Junior ) = 0.76596 approximately

4 0
3 years ago
A second important result is that electrons will fill the lowest energy states available. This would seem to indicate that every
Assoli18 [71]

Answer:

  • 8

Explanation:

1)<u> Principal quantum number, n = 2</u>

  • n is the principal quantum number and indicates the main energy level.

<u>2) Second quantum number, ℓ</u>

  • The second quantum number, ℓ,  is named, Azimuthal quantum number.

The possible values of ℓ are from 0 to n - 1.

Hence, since n = 2, there are two possible values for ℓ: 0, and 1.

This gives you two shapes for the orbitals: 0 corresponds to "s" orbitals, and 1 corresponds to "p" orbitals.

<u>3) Third quantum number, mℓ</u>

  •    The third quantum number, mℓ, is named magnetic quantum number.

The possible values for mℓ are from - ℓ to + ℓ.

Hence, the poosible values for mℓ when n = 2 are:

  • for ℓ = 0: mℓ = 0
  • for ℓ = 1, mℓ = -1, 0, or +1.

<u>4) Fourth quantum number, ms.</u>

  • This is the spin number and it can be either +1/2  or -1/2.

Therfore the full set of possible states (different quantum number for a given atom) for n = 2 is:

  • (2, 0, 0 +1/2)
  • (2, 0, 0, -1/2)
  • (2, 1, - 1, + 1/2)
  • (2, 1, -1, -1/2)
  • (2, 1, 0, +1/2)
  • (2, 1, 0, -1/2)
  • (2, 1, 1, +1/2)
  • (2, 1, 1, -1/2)

That is a total of <u>8 different possible states</u>, which is the answer for the question.

8 0
3 years ago
Other questions:
  • 20% of a restaurants customers order the chefs special. 230 customers are anticipated to dine at the restaurant tonight. The sta
    5·1 answer
  • I need help please!!
    7·1 answer
  • Need help ASAP! Simplify the expression so there is only one positive power for the base, -5. (The questions/answers are in the
    9·2 answers
  • (-4)(9)+(-55 divided -5)=
    5·1 answer
  • The chart shows the number of points Jerry scored on each test this week out of the number of points possible for the test.
    7·1 answer
  • Compare and contrast the methods you can use to multiply a whole number and a decimal.
    12·1 answer
  • The volume of a cone is 300cm3 and the height is 10cm3
    11·1 answer
  • Somebody please help me out.
    14·1 answer
  • Help i dont know how to do this type of questions​
    11·2 answers
  • choosing a card from a deck of cards numbered 10, 11, 12, ..., 25 and picking a day of the week WHAT IS THE PROBABILITY
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!