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3 years ago

Which function is represented by the graph below?

Mathematics

2 answers:

strojnjashka [21]3 years ago

8 0

Answer:

Equation: $f(x)=3^x$

A is correct

Step-by-step explanation:

Given: The graph of a function on x-y coordinate plane.

y-intercept: (0,1)

Graph starts from II quadrant touches x-axis at infinity.

End in I quadrant increases gradually.

This is property of exponential function.

$y=ab^x$

y-intercept: (0,1)

$1=a$

Passing point: (1,3)

$3=b^1$

b=3

Equation of graph:

$y=3^x$

Hence, The equation of given graph is $y=3^x$

Zina [86]3 years ago

5 0

This would be an exponential graph and non of the answers are exponential. but the graph would be f(x)=3^x

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Answer:

(B) A selected group of participants that is representative of a larger population

Step-by-step explanation:

A sample statistic is a piece of information that you get from a fraction of the population. This information is representative of a larger population.

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Answer:

Step-by-step explanation:

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2 years ago

Which of the following equations is modeled by the graph.

Sergio [31]

Answer:

a = 50t

Step-by-step explanation:

The graph goes through (0,0) so the y intercept is 0

a=5t

a=50t

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The slope is y/x since the y intercept is 0

m = 50/1 = 50

The equation for the line is

y = 50x+0

y = 50x

Using the variables given

a = 50t

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3 years ago

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Answer:

Step-by-step explanation:

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2 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$

Therefore the half-life is between 600 and 700 years

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2 years ago