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vovikov84 [41]
3 years ago
6

Does the table represent an exponential function?

Mathematics
2 answers:
serious [3.7K]3 years ago
6 0
Yes it does represent an exponential function

if you put the table into a graphing calculator it will show as an exponential function

desmos is a free one online
NeX [460]3 years ago
4 0

No, the table represents an exponential function. The function is \boxed{f\left( x \right) =  - {{\left( x \right)}^3}}.

Further explanation:

The linear equation with slope m and intercept c is given as follows.

\boxed{y = mx + c}

The standard form of the linear equation can be expressed as follows,

\boxed{ax + by = c}

Given:

The values of x are 1, 2, 3 and 4.

The values of y are -1, -8, -27 and -64.

Explanation:

The general form of exponential function can be expressed as follows,

f\left( x \right) = a{b^x}

The points are \left( {1, - 1} \right),{\text{ }}\left( {2, - 8} \right),{\text{ }}\left( {3, - 27} \right){\text{ and }}\left( {4, - 64} \right).

The function for the given points can be expressed as follows,

f\left( x \right) = - {\left( x \right)^3}

No the table represents an exponential function. The function is \boxed{f\left( x \right) = - {{\left( x \right)}^3}}.

Learn more:

  1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.
  2. Learn more about equation of circle brainly.com/question/1506955.
  3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Exponentialfunction

Keywords: numbers, standard form, point slope form, exponential function,slope, slope intercept, inequality, equation.

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Answer:

(a) The value of P (X = 7) is 0.1388.

(b) The value of P (X ≥ 3) is 0.9380.

(c) The value of P (2 < X < 7) is 0.5433.

(d) \mu_{X}=6

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Step-by-step explanation:

Let <em>X</em> = number of uranium fission tracks on per cm² surface area of the mineral.

The average number of track per cm² surface area is, <em>λ</em> = 6.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 6.

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(a)

Compute the value of P (X = 7) as follows:

P(X=6)=\frac{e^{-6}(6)^{7}}{7!}=\frac{0.0025\times 279936}{5040}=0.1388

Thus, the value of P (X = 7) is 0.1388.

(b)

Compute the value of P (X ≥ 3) as follows:

P (X ≥ 3) = 1 - P (X < 3)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2)

              =1-\frac{e^{-6}(6)^{0}}{0!}-\frac{e^{-6}(6)^{1}}{1!}-\frac{e^{-6}(6)^{2}}{2!}\\=1-0.00248-0.01487-0.04462\\=0.93803\\\approx0.9380

Thus, the value of P (X ≥ 3) is 0.9380.

(c)

Compute the value of P (2 < X < 7) as follows:

P (2 < X < 7) = P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)

                   =\frac{e^{-6}(6)^{3}}{3!}+\frac{e^{-6}(6)^{4}}{4!}+\frac{e^{-6}(6)^{5}}{5!}+\frac{e^{-6}(6)^{6}}{6!}\\=0.08924+0.13385+0.16062+0.16062\\=0.54433\\\approx0.5443

Thus, the value of P (2 < X < 7) is 0.5433.

(d)

The mean of the Poisson distribution is:

\mu_{X}=\lambda=6

(e)

The standard deviation of the Poisson distribution is:

\sigma_{X}=\sqrt{\sigma^{2}_{X}}=\sqrt{\lambda}=\sqrt{6}=2.4495\approx2.45

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