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4 years ago

Which is a solution of x2-x-3/4=0

Mathematics

1 answer:

SIZIF [17.4K]4 years ago

6 0

If you're trying to figure out what x is equal to, it's 3/2, or -1/2

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Please help me asap

Elenna [48]

Answer:

Not sure what number sense is, but 3 is the answer.

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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

3 years ago

What set of transformations is performed on ABCD to form A′B′C′D′?

Ilia_Sergeevich [38]

Reflection across the y-axis, one unit down, one unit to the right.

Hope It Helps!

7 0

3 years ago

Which value for x makes the equation x -4.21 = 6.047 true? A. x = 10.68 C. X = 10.247 B. x = 10.257 D. x = 1.837

Anuta_ua [19.1K]

Answer:

C. X = 10.247

Step-by-step explanation:

X -4.21 = 6.047

just you will add 4.21 to both sides of the equation

X -4.21 + 4.21 = 6.047 + 4.21

X = 10.247

3 0

2 years ago

Read 2 more answers

There are 68 apples in the school kitchen. The school needs 130 apples for all the students in grade 2. The school gets two boxe

WITCHER [35]

Answer:No they do not have enough they only have 82 apples

Step-by-step explanation:

7 0

3 years ago