All categories

3 years ago

Why do we use variables, and what is the significance of using them in expressions and equations?

1 answer:

5 0

Variables are letters from the end of the alphabet
Economics always uses the firsts letter of whatever that letter is representing.

 An expression such as 4x3 is a variable. It can assume different values because x can assume different values.

You might be interested in

Learning Thoery In a learning theory project, the proportion P of correct responses after n trials can be modeled by p = 0.83/(1

elena-s [515]

Answer:

a) $P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

b) $P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

c) $0.75 =\frac{0.83}{1+e^{-0.2n}}$

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

d) If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

Step-by-step explanation:

For this case we have the following expression for the proportion of correct responses after n trials:

$P(n) = \frac{0.83}{1+e^{-0.2t}}$

Part a

For this case we just need to replace the value of n=3 in order to see what we got:

$P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536$

So the number of correct reponses after 3 trials is approximately 0.536.

Part b

For this case we just need to replace the value of n=7 in order to see what we got:

$P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666$

So the number of correct responses after 7 weeks is approximately 0.666.

Part c

For this case we want to solve the following equation:

$0.75 =\frac{0.83}{1+e^{-0.2n}}$

And we can rewrite this expression like this:

$1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}$

$e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}$

Now we can apply natural log on both sides and we got:

$ln e^{-0.2n} = ln (\frac{8}{75})$

$-0.2 n = ln(\frac{8}{75})$

And then if we solve for t we got:

$n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials$

And we can see this on the plot attached.

Part d

If we find the limit when n tend to infinity for the function we have this:

$lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83$

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

5 0

2 years ago

Help pls ssssssssssssssssssssssssssssss

kirza4 [7]

Answer:

The numbers , -0.82 , -1/5 , and -4/5 are closest to - 1

The number, -0.28 is the ONLY number closest to - 1/4

hope this helps

8 0

2 years ago

Teresa is building a scale model of the JP Morgan chase Tower in Houston.In her scale model, 1.5 inches represents 61 meters. Si

Jet001 [13]

Answer:

Teresa should build 7.5 inches model.

Step-by-step explanation:

Consider the provided information.

Teresa is building a scale model of the JP Morgan chase Tower in Houston.

Her scale model, 1.5 inches represents 61 meters.

The actual building is 305 meters tall,

First find how many 61's are there in 305.

For this divide 305 by 61.

$\frac{305}{61}= 5$

Actual building is 5 times taller than 61. Thus the model should be 5 times taller than 1.5.

multiply 1.5 with 5 gives us:

$1.5\times 5=7.5$

Hence, Teresa should build 7.5 inches model.

7 0

2 years ago

Need help with 5 6 7 and 8. at least answer one of them .

likoan [24]

Answer:

5) moves one place to right

6) moves one place to left

Step-by-step explanation:

8 0

3 years ago

Damian spends a winter day recording the temperature once every three hours for science class. At 9 am, the temperature was -6°F

Jlenok [28]

Answer:

-4.2

Step-by-step explanation:

6.9 - 11.1

8 0

3 years ago