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MrRa [10]
3 years ago
13

Which graph models the function f(x)=-2(3)^x

Mathematics
2 answers:
Vitek1552 [10]3 years ago
8 0

Answer:

Reflection across x-axis:

(x, y) \rightarrow (x, -y)

Vertically stretch:

A function y=a f(x) is vertically stretch by a factor a > 1 is that of parent function y=f(x)

Given the graph: f(x) = -2(3)^x

We will make a table values for a few values of x.

then we will graph the given function.

x                f(x)

-2            -0.2222..

-1             -0.6666..

0                 -2

1                  -6

2                 -18

3                 -54

Note that as x increases, f(x) decreases

Now, using these points (-2, -0.2222..), (-1, -0.666..), (0, -2), (1, -6) , (2, -18) and (3, -54)

Plot the graph of the given function as shown below:

We observe that the curve is that of f(x) =(3)^x except it is vertically stretch by a factor 2 and reflection across x-axis.

svetlana [45]3 years ago
4 0

Solution:

\text{Given: }f(x)=-2(3)^x

We are an exponential function.

y=ab^x

If a<0 and b>1 then graph is decreasing.

If a>0 and b<1 then graph is decreasing.

If a<0 and b<1 then graph is increasing.

If a>0 and b>1 then graph is increasing.

\text{We have a function }f(x)=-2(3)^x

Here a=-2<0 and b=3>1 therefore, f(x) is decreasing.

Horizontal asymptote , y=0

x-intercept: Doesn't not exist

y-intercept: (0,-2)

Using above information we will draw the graph f(x)

We make table of x and y for different value of x

  x        y

-2       -0.22

 -1       -0.67

 0          -2

 1           -6

 2           -18

Plot these points on graph and join the points.

 Please see the attachment to see the graph.



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B) Unusually small

C) process was no longer functioning correctly

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E) Not unusually large

F) No-Evidence

G) P ( X ≤ 195 ) = 0.3085

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I) No evidence

Step-by-step explanation:

Given:-

- The random variable (X) denotes the adhesive strength is normally distributed with mean u = 200 N and standard deviation s.d = 10 N.

                            X ~ N ( 200 , 10^2 )

Solution:-

A) Find P(X ≤ 160), under the assumption that the process is functioning correctly.

- Determine the Z-score value:

                      Z-score = ( X - u ) / s.d

                                    = ( 160 - 200 ) / 10

                                    = -4

- Use the standardized z-table to determine the probability:

                      P ( X ≤ 160 ) = P ( Z ≤ -4 )

                                           = 0

- Assuming the process is functioning properly then the adhesive strength of X = 160 N would be considered unusually small since the probability of occurrence is approximately 0.

- If we were to observe an adhesive strength process that gives us the value of 160 N can imply that the process is not functioning properly as its outside the 3 standard deviations from the mean value. ( Conclusive Evidence )

D) Find P(X ≥ 203), under the assumption that the process is functioning correctly.

- Determine the Z-score value:

                      Z-score = ( X - u ) / s.d

                                    = ( 203 - 200 ) / 10

                                    = 0.3

- Use the standardized z-table to determine the probability:

                      P ( X ≥ 203 ) = P ( Z ≥ 0.3 )

                                           = 0.3821

- Assuming the process is functioning properly then the adhesive strength of X = 203 N would be not be considered unusually small since the probability of occurrence is in the heart of the bell curve.

- If we were to observe an adhesive strength process that gives us the value of 203 N can imply that the process is functioning properly as its within 1 standard deviation from the mean value. ( No evidence )

G) Find P(X ≤ 195), under the assumption that the process is functioning correctly.

- Determine the Z-score value:

                      Z-score = ( X - u ) / s.d

                                    = ( 195 - 200 ) / 10

                                    = -0.5

- Use the standardized z-table to determine the probability:

                      P ( X ≤ 195 ) = P ( Z ≤ -0.5 )

                                           = 0.3085

- Assuming the process is functioning properly then the adhesive strength of X = 195 N would be not be considered unusually small since the probability of occurrence is in the heart of the bell curve.

- If we were to observe an adhesive strength process that gives us the value of 195 N can imply that the process is functioning properly as its within 1 standard deviation from the mean value. ( No evidence )

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