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3 years ago

Find all real solutions to w^2-10w+1 In simplest form

Mathematics

1 answer:

Anvisha [2.4K]3 years ago

6 0

Answer:

you can solve it using quadratic formula

w=5-2 √6 or w=5+ 2√6

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1 + 1 - 1 = ? For fun- .........

coldgirl [10]

Answer:

1+1=2-1=1

Step-by-step explanation:

lol, this is the hardest question I've done today.

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2 years ago

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B + 1.5 = 6 solve the equation and find the value of the variable

pickupchik [31]

The answer is 4.5. Because 6 minus 1.5 is equal to 4.5

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2 years ago

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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

Write the ordered pair that represents YZ. Then find the magnitude of YZ. y(-4,12), Z(1,19).

ki77a [65]

The magnitude of YZ is 8.6

Explanation:

Y( -4, 12)

Z ( 1, 19)

Magnitude of YZ = ?

We know:

$YZ = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

On substituting the value we get:

$YZ = \sqrt{(1+4)^2 + (19 - 12)^2} \\\\YZ = \sqrt{(5)^2 + (7)^2} \\\\YZ = \sqrt{25+49} \\\\YZ = \sqrt{74} \\\\YZ = 8.6$

Thus, the magnitude of YZ is 8.6

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3 years ago

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Find the area of a circle if the radius is 2.5m. Round your answer to the nearest hundredth

abruzzese [7]

Answer: 19.63

Step-by-step explanation:

A = pir^2 , pi*2.5^2 = 19.63

5 0

3 years ago