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Finger [1]
3 years ago
7

HELP HELP HELP

Mathematics
2 answers:
mojhsa [17]3 years ago
8 0

Answer:

2998 ; 17%

Step-by-step explanation:

Given the function:

t(c)=-3970.9(ln c)

c = % of carbon remaining ; t = time

1.) c = 47% = 47/100 = 0.47

t(0.47) = - 3970.9(In 0.47)

t = - 3970.9 * −0.755022

t = 2998.119

t = 2998

B.)

t = 7000

t(c)=-3970.9(ln c)

7000 = - 3970.9(In c)

7000 / - 3970.9 = In c

−1.762824 = In c

c = exp(−1.762824)

c = 0.1715596

c = 0.1715596 * 100%

c = 17.156% ; c = 17%

Lapatulllka [165]3 years ago
5 0

Answer:  For plato family

The function t relates the age of a plant fossil, in years, to the percentage, c, of carbon-14 remaining in the fossil relative to when it was alive.

Use function t to complete the statements.

A fossil with 47%    of its carbon-14 remaining is approximately

years old.

A fossil that is 7,000 years old will have approximately

       17%     of its   carbon-14 remaining.

Step-by-step explanation:

 The  first  box is  2,998

  The  second  box is 17%

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Substitute for y and rewrite the system of equations as a single equation which can be used to solve for x.
Vlada [557]

Answer:

x = 3  and y = 1

Step-by-step explanation:

Given equations

y=3x-8 and y=4-x

We will substitute y from second equation in the first and get:

4-x = 3x-8   first we will add (+x) to both sides and get

4-x+x = 3x+x-8 =>   4 = 4x-8 now we will add (+8) to the both sides and get

4+8 = 4x-8+8 => 12 = 4x  now we will change the sides of equation and get

4x = 12 => x = 12/4 = 3 => x=3  now we will replace x=3 in the second equation

y= 4-x = 4-3=1 => y=1  

Finally we have  (x , y) = ( 3 , 1)

God with you!!!

7 0
3 years ago
the cost of a candle varies directly with the volume of the candle. If a candle with a volume of 80cm^3 costs $12, what would be
ivanzaharov [21]
<span>First, you need to find out how much will the candle cost per volume. If a candle with a volume of 80cm^3 costs $12, the price of the candle per cm^3 would be: $12/80cm^3= $0.15/cm^3

Then the price of a candle with </span>120 cm^3 volume would be:120cm^3* $0.15/cm^3= $18
4 0
3 years ago
In a standard set of dominoes, a face of each domino has a line through the center, with 0 to 6 dots on each side of the line. E
beks73 [17]
1/7 of the dominoes will have the same number on both sides
5 0
3 years ago
The Midpoint of AB dkdldl
Nutka1998 [239]
  • The Midpoint of AB is (1,0).

Given that:

  • In line AB, where the coordinates of A is (3,1) and coordinates of B is (-1,-1).

To find:

  • The Midpoint of line AB.

So, according the question

We know that,

The midpoint M of a line segment AB with endpoints A (x₁, y₁) and B (x₂, y₂) has the coordinates M ( \frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} ).

Now from question,

We know that the the coordinates of A is (3,1) and coordinates of B is (-1,-1) of line AB.

So, we can say that

   A is (3,1) or x₁ = 3 and y₁ = 1.

   B (-1,-1) or x₂ = -1 and y₂ = -1.

∵ The coordinates of midpoint M (X,Y)

          X = \frac{x_1 + x_2}{2}

              = \frac{3-1}{2}

              = 2/2

          X = 1.

And

          Y = \frac{y_1 + y_2}{2}

              = \frac{1-1}{2}

              = 0/2

          Y = 0.

So, the midpoint of line AB is M (1,0)

To learn more about Midpoint of line, please click on the link;

brainly.com/question/14687140

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8 0
2 years ago
The probability of tossing heads with a standard coin is 1/2, because it is one of two possible outcomes. The probability of tos
spayn [35]

Answer:

  a) 1/64

  b) 1/4096

Step-by-step explanation:

As you can tell from the example, the exponent of 1/2 is the number of heads in a row.

a) p(6 heads in a row) = (1/2)^6 = 1/(2^6) = 1/64

b) p(12 heads in a row) = (1/2)^12 = 1/(2^12) = 1/4096

_____

<em>Additional comment</em>

The probability of a head is 1/2 because we generally are concerned with a "fair coin." That is defined as a coin in which each of the 2 possible outcomes has the same probability, 1/2. Similarly, a "fair number cube" has 6 faces, and the probability of each is defined to be the same as any other, 1/6. Loaded dice and unfair coins do sometimes show up in probability problems.

3 0
2 years ago
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