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kobusy [5.1K]
3 years ago
14

Which equations represent exponential growth? Which equations represent exponential decay? Drag the choices into the boxes to co

mplete the table.

Mathematics
1 answer:
Norma-Jean [14]3 years ago
5 0

Answer:

Exponential growth functions are:

A=20000(1.08)^{t}

A=40(3)^{t}

A=1700(1.07)^{t}

Exponential decay functions are:

A=80(\frac{1}{2})^{t}=80(0.5)^{t}

A=1600(0.8)^{t}

A=1700(0.93)^{t}

Step-by-step explanation:

Given:

An exponential function is of the form y=ab^{x}, where, a\ne 0.

Now, if a > 0 and b > 1, then the exponential function represent exponential growth.

If a > 0 and 0 < b < 1, then the exponential function represent exponential  decay.

Let us check each function now.

Option 1: A=20000(1.08)^{t}

Here, a = 20000, b = 1.08

As 1.08 > 1, the function is exponential growth.

Option 2: A=80(\frac{1}{2})^{t}=80(0.5)^{t}

Here, a = 80, b = 0.50

As 0.5 < 1, the function is exponential decay.

Option 3: A=1600(0.8)^{t}

Here, a = 1600, b = 0.8

As 0.8 < 1, the function is exponential decay.

Option 4: A=40(3)^{t}

Here, a = 40, b = 3

As 3 > 1, the function is exponential growth.

Option 5: A=1700(1.07)^{t}

Here, a = 1700, b = 1.07

As 1.07 > 1, the function is exponential growth.

Option 6: A=1700(0.93)^{t}

Here, a = 1700, b = 0.93

As 0.93 < 1, the function is exponential decay.

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oksano4ka [1.4K]

Yes it is true:

(1 - cos^2 x) = sin^2 x Sqr.(1 - cos^2 x) = sin x and (-sinx) True, since sin x negative. Also, x terminates in quadrant 3 and 4. Hope this is helpful

4 0
2 years ago
Daily high temperatures in St. Louis for the last week were as​ follows: 95​, 92​, 93​, 92​, 95​, 90​, 90 ​(yesterday). ​a) The
hichkok12 [17]

Answer:

a) T = 91.7 degrees

b) T = 90 degrees

c) MAD = 1.9

d) MSE = 5.05

Step-by-step explanation:

Given:

- Daily high temperatures in St. Louis for the last week were as​ follows:

                                   95​, 92​, 93​, 92​, 95​, 90​, 90

Find:

a) Forecast the high temperature today, using a 3-day moving average.

b) Forecast the high temperature today, using a 2-day moving average.

c) Calculate the mean absolute deviation based on a 2-day moving average, covering all days in which you can have a forecast and an actual temperature.

d) The mean squared error for the​ 2-day moving average​

Solution:

a)

- The set of 3 day moving average is as follows:

4.   (95 + 92 + 93) ÷ 3 = 93.33⁰C

5.  (92 + 93 + 92) ÷ 3 = 92.33⁰C

6.  (93 + 92 + 95) ÷ 3 = 93.33⁰C

7.  (92 + 95 + 90) ÷ 3 = 92.33⁰C

8.  (95 + 90 + 90) ÷ 3 = 91.667⁰C

- Now use these points on excel sheet to forecast the temperature for today. The line of best fit is given:

T = 91.7 degrees

b)

- The set of 2 day moving average is as follows:

3.   (95 + 92) ÷ 2 = 93.5⁰C

4.  (95 + 93) ÷ 2 = 92.5⁰C

5.  (93 + 92) ÷ 2 = 92.5⁰C

6.  (92 + 95) ÷ 2 = 93.5⁰C

7.  (95 + 90) ÷ 2 = 92.5⁰C

8. (90 + 90) ÷ 2 = 90⁰C

- Now use these points on excel sheet to forecast the temperature for today. The line of best fit is given:

T = 90 degrees

c)

                             Error             Error^2

3.   93.5⁰C            0.5                  0.25

4.   92.5⁰C            0.5                  0.25

5.   92.5⁰C            2.5                  6.25

6.   93.5⁰C            3.5                  12.25

7.   92.5⁰C            2.5                  6.25

8.   90⁰C

- The mean absolute deviation as follows:

                              MAD = Sum of all errors  / 5

                              MAD = (0.5+0.5+2.5+3.5+2.5)  / 5

                              MAD = 1.9

d)

- The mean squared error deviation as follows:

                              MSE = Sum of all error^2  / 5

                              MSE = (0.25+0.25+6.25+12.25+6.25)  / 5

                              MSE = 5.05

4 0
3 years ago
An urn contains nine red and one blue balls. A second urn contains one red and five blue balls. One ball is removed from each ur
il63 [147K]

Answer:

0.362

Step-by-step explanation:

When drawing randomly from the 1st and 2nd urn, 4 case scenarios may happen:

- Red ball is drawn from the 1st urn with a probability of 9/10, red ball is drawn from the 2st urn with a probability of 1/6. The probability of this case to happen is (9/10)*(1/6) = 9/60 = 3/20 or 0.15. The probability that a ball drawn randomly from the third urn is blue given this scenario is (1 blue + 5 blue)/(8 red + 1 blue + 5 blue) = 6/14 = 3/7.

- Red ball is drawn from the 1st urn with a probability of 9/10, blue ball is drawn from the 2nd urn with a probability of 5/6. The probability of this event to happen is (9/10)*(5/6) = 45/60 = 3/4 or 0.75. The probability that a ball drawn randomly from the third urn is blue given this scenario is (1 blue + 4 blue)/(8 red + 1 blue + 1 red + 4 blue) = 5/14

- Blue ball is drawn from the 1st urn with a probability of 1/10, blue ball is drawn from the 2nd urn with a probability of 5/6. The probability of this event to happen is (1/10)*(5/6) = 5/60 = 1/12. The probability that a ball drawn randomly from the third urn is blue given this scenario is (4 blue)/(9 red + 1 red + 4 blue) = 4/14 = 2/7

- Blue ball is drawn from the 1st urn with a probability of 1/10, red ball is drawn from the 2st urn with a probability of 1/6. The probability of this event to happen is (1/10)*(1/6) = 1/60. The probability that a ball drawn randomly from the third urn is blue given this scenario is (5 blue)/(9 red + 5 blue) = 5/14.

Overall, the total probability that a ball drawn randomly from the third urn is blue is the sum of product of each scenario to happen with their respective given probability

P = 0.15(3/7) + 0.75(5/14) + (1/12)*(2/7) + (1/60)*(5/14) = 0.362

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Answer:

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