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3 years ago

How many year apart is 384 BC from 1952 AD

2 answers:

6 0

BC and BCE are Before Common Era
So what you want to do is add 1952 and 384
1952 + 384 = 2336
So the amount of years between 384 BC and 1952 AD is about 2,336 years
Hope this helps

Ne4ueva [31]3 years ago

4 0

So you have to make a number line from 384 bc a negative numb r to 1952 ad this is a positive number for example -384+1952= then you have your answer

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Answer:

-3 is the answer for the question

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lauren is saving for gymnastics camp camp cost 225 attend she got 40 percent of the money saved how much money has she saved

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She has saved 90 dollars

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3 years ago

Suppose the polynomial f(x) has the following end behavior: as x approaches infinity, f(x) approaches infinity, and as x approac

galina1969 [7]

Answer:

C. x³+10x²−5x+5

E. 7x⁵+4x²

F. x+8

Step-by-step explanation:

The graph they described should look something like the one i drew below.

So all you have to do is look at the number with the highest exponent.

If the exponent is an odd number and the x is positive then that function will have an end behavior like the picture i posted.

x³ is good

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3 years ago

P/9 - 1 < -2 ... Solve for p

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Step-by-step explanation:

first collect terms, while the < remains unchange

P < 1-2

cross examine and get the answer

P < -1

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7 0

2 years ago

7. In a state lottery, a player must choose 8 of the numbers from 1 to 40. The lottery commission then performs an experiment th

Vlad1618 [11]

Answer:a) P(8 of the players numbers are drawn)=1.3×10^-8

b) P(7 of the players number are drrawn)=3.33×10^-c) P(at least 6 of the players number were drawn)=1.84×10^-4

Step-by-step explanation:

Players has 8 combinations of numbers from 1-40. The outcome S contains all the combinations of 8 out of 40

a) P(8 of the players numbers are drawn)= 1/40/8= 1.3×10^-8

There are one in hundred million chances that the draw numbers are precisely the chosen ones.

b) Number of ways of drawing 78 selected numbers from 1-40=8×(40-7)

8×32

P(7 of the players number are drawn)=8×32/40 =3.33×10^-6.

There are approximately 300,000 chances that 7 of the players numbers are chosen

c) P(at least 6 players numbers are drawn)= 32/2×(8/6) ways to draw.

P(at least 6 players numbers are drawn)=P(all 8 chosen are drawn)+P(7 players numbers drawn)+P(6 chosen are drawn) = 1+ 8 x32/40/8 +[8\6 ×32/2]

P(at least 6 players numbers are drawn) = 1.84×10^-4.

There are approximately 5400chances that at least6 of the numbers drawn are chosen by the player.

5 0

3 years ago