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Vinil7 [7]
3 years ago
11

Please help need it

Mathematics
2 answers:
Rudiy273 years ago
6 0
It should be infinity.
elixir [45]3 years ago
5 0
THe answer is infinity
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A scientist measured the energy of a system and found that it was increasing. The data shown represent the energy taken every bi
DiKsa [7]

At start ( t = 0 nanoseconds ) :

E ( t = 0 ) = 2.645 J

E ( t = 1 ) = 6.290 J

E ( t = 1 ) : E ( t = 0 ) = 6.290 : 2.645 = 2.37

Also:

E ( t = 2 ) : E ( t = 1 ) = 14.909 : 6.290 = 2.37

E ( t = 3 ) ; E ( t = 2 ) = 35.335 : 14.909 = 2.37

Therefore, the formula for calculating the energy of the system is:

E ( t ) = 2.645 * 2.37 ^ t

The answer is 4) exponential growth.


3 0
3 years ago
Read 2 more answers
What is: ⅛ ( 64v + 24) =
Illusion [34]

Answer:

8v + 3.

Step-by-step explanation:

1/8 * 64v  + 1/8 * 24

4 0
2 years ago
Read 2 more answers
I’m on a test so I need help
sladkih [1.3K]
The answer is 2 y2 -9
4 0
3 years ago
Solve the equation.<br><br>-4 (y - 2) = 12<br><br>y = ?​
inna [77]
First you distribute what’s outside the parentheses (the -4)

-4•4= -4y
-4•-2= 8

New equation: -4y+8=12

Now you solve like a normal equation
12-8 is 4

New equation: -4y=4

Now divide

Answer:-1
8 0
2 years ago
Scott and Letitia are brother and sister. After dinner, they have to do the dishes, with one washing and the other drying. They
ehidna [41]

Answer:

The probability that Scott will wash is 2.5

Step-by-step explanation:

Given

Let the events be: P = Purple and G = Green

P = 2

G = 3

Required

The probability of Scott washing the dishes

If Scott washes the dishes, then it means he picks two spoons of the same color handle.

So, we have to calculate the probability of picking the same handle. i.e.

P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)

This gives:

P(G_1\ and\ G_2) = P(G_1) * P(G_2)

P(G_1\ and\ G_2) = \frac{n(G)}{Total} * \frac{n(G)-1}{Total - 1}

P(G_1\ and\ G_2) = \frac{3}{5} * \frac{3-1}{5- 1}

P(G_1\ and\ G_2) = \frac{3}{5} * \frac{2}{4}

P(G_1\ and\ G_2) = \frac{3}{10}

P(P_1\ and\ P_2) = P(P_1) * P(P_2)

P(P_1\ and\ P_2) = \frac{n(P)}{Total} * \frac{n(P)-1}{Total - 1}

P(P_1\ and\ P_2) = \frac{2}{5} * \frac{2-1}{5- 1}

P(P_1\ and\ P_2) = \frac{2}{5} * \frac{1}{4}

P(P_1\ and\ P_2) = \frac{1}{10}

<em>Note that: 1 is subtracted because it is a probability without replacement</em>

So, we have:

P(Same) = P(G_1\ and\ G_2) + P(P_1\ and\ P_2)

P(Same) = \frac{3}{10} + \frac{1}{10}

P(Same) = \frac{3+1}{10}

P(Same) = \frac{4}{10}

P(Same) = \frac{2}{5}

8 0
2 years ago
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