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NARA [144]
3 years ago
12

Please help with question 9

Mathematics
1 answer:
Dmitrij [34]3 years ago
4 0

Answer:

y=x + 6

Step-by-step explanation:

yeah-ya........ right?

You might be interested in
Eight rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other ro
Brums [2.3K]

Answer:

9.11*10^-4 %

Step-by-step explanation:

To find the probability, you simply need to find the possible outcomes that allows no rooks to be in danger, and the possible amount of ways to place the rooks.

For the first outcome, you start by putting 1 rook in the first columns, you have 8 possible rows to do this. The next rook in the next column will only have 7 possible rows, as you have to exclude the one where the previous rook is located. The next rook, 6 possibilities, the next 5, and so on. So we conclude that the total amount of ways so that none of the rooks can capture any of the other rooks is 8*7*6*5*4*3*2*1 = 8! = 40320

In order to find the total amount of ways to place the rooks, you can just use a combinatoric:

\left[\begin{array}{ccc}64\\8\end{array}\right]= \frac{64!}{8!(64-8)!} = 4.43*10^9

Then:

P = \frac{40320}{4.43*10^9}*100\%=9.11*10^{-4}\%

5 0
4 years ago
The perimeter of a square garden is 54 feet. What is the length of one side of the garden?
Dahasolnce [82]

Answer: 13.5

Step-by-step explanation:

54/4 = 13.5

13.5 x 4 = 54

4 0
3 years ago
A farmer in China discovers a mammal hide that contains 37% of its original amount of C-14. Find the age of the mammal hide to t
Naya [18.7K]

Answer: 54678 years

Step-by-step explanation:

This can be solved by the following equation:

N_{t}=N_{o}e^{-\lambda t} (1)

Where:

N_{t}=54\%=0.54 is the quantity of atoms of carbon-14 left after time t

N_{o}=1 is the initial quantity of atoms of C-14 in the mammal hide

\lambda is the rate constant for carbon-14 radioactive decay

t is the time elapsed

On the other hand, \lambda has a relation with the half life h of the C-14, which is 5730 years:

\lambda=\frac{ln(2)}{h}=\frac{ln(2)}{5730 years}=1.21(10)^{-4} years^{-1}=0.000121 years^{-1} (2)

Substituting (2) in (1):

0.54=1e^{-(0.000121 years^{-1}) t} (3)

Applying natural logarithm on both sides of the equation:

ln(0.54)=ln(1e^{-(0.000121 years^{-1}) t}) (4)

-0.616=-(0.000121 years^{-1}) t (5)

Isolating t:

t=\frac{-0.616}{-0.000121 years^{-1}} (6)

t=54677.68 years \approx 54678 years (7)  This is the age of the mammal hide

8 0
3 years ago
The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there a
ad-work [718]

Answer: 3,917 days

Step-by-step explanation:

5 0
3 years ago
What happens to the volume of a rectangular prism when each of the dimensions is doubled?
bonufazy [111]
D is the answer........
8 0
4 years ago
Read 2 more answers
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