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romanna [79]
4 years ago
11

Does each equation represent exponential decay or exponential growth? Drag and drop the choices into the boxes to correctly comp

lete the table. Note: If an equation is neither exponential growth nor exponential decay, do not drag it to the table.
W=9/8(3/5)t
L=0.25(12)t
f(t)=(3/4)t
W=0.5(2.1)t
L=4.2(0.6)t
f(t)=2/3(6)t
g(x)=1.3(x)
Mathematics
1 answer:
konstantin123 [22]4 years ago
3 0

in plain and short, if the number inside the parentheses is less than 1, that's a flag is a decay, if it's more than 1, is growth.

3/5 is less than 1

12 is clearly more than 1

3/4 is less than 1

2.1 is more than 1

0.6 is less than 1

6 is more than 1.

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Suppose that IQs of East State University’s students can be described by a Normal model with mean 130 and standard deviation 8 p
lianna [129]

Answer:

a) P(X \geq 125) = 1-P(X

b) P(R \geq 5) = 1-P(R

c) P(\bar Y \geq 125) = P(Z> \frac{125-120}{5.774}) = 1-P(Z

d) P(H\geq 5)= P(Z> \frac{5-10}{7.394}) = 1-P(Z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the scores for the East population, and for this case we know the distribution for X is given by:

X \sim N(130,8)  

Where \mu_x=130 and \sigma_x=8

Let Y the random variable that represent the scores for the West population, and for this case we know the distribution for Y is given by:

X \sim N(120,10)  

Where \mu_y=120 and \sigma_y=10

Part a

For this case we want this probability:

P(X\geq 125)

And we can use the z score given by:

z = \frac{X -\mu}{\sigma}

And if we replace we got:

P(X \geq 125) = 1-P(X

Part b

For this case we need to define the following random variable R = X-Y and we know that the distribution of R is given by:

R \sim N (130-120= 10, \sigma_R = \sqrt{8^2 +10^2}=12.806)

And we want this probability:

P(R\geq 5)

We can use the z score given by:

z= \frac{R -\mu_R}{\sigma_R}

If we use this formula we got:

P(R \geq 5) = 1-P(R

Part c

For this case we select a sample size of n =3 for the Y distribution, the sample mean have the following distribution:

\bar Y \sim N(120, \frac{10}{\sqrt{3}}=5.774)

And we want this probability:

P(\bar Y \geq 125) = P(Z> \frac{125-120}{5.774}) = 1-P(Z

Part d

For this case we define the following random variable H = \bar X -\bar Y and the distribution for H is given by:

H \sim N (130-120=10, \sigma_H = \sqrt{\frac{8^2 +10^2}{3}}= 7.394)

And the z score would be given by:

z = \frac{H -\mu_H}{\sigma_H}

And if we find the probability required we got:

P(H\geq 5)= P(Z> \frac{5-10}{7.394}) = 1-P(Z

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In the given graph, the two lines intersects at

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f(x) = g(x)

And at x=-2, the y values are equal. THat is

f(-2) = g(-2)

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