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2 years ago

HELP ASAP ILL GIVE BRAINLIST

Mathematics

1 answer:

Mila [183]2 years ago

5 0

Answer:

Part A) ΔCGB and ΔPBE are similiar

Part B) They are similar because they are Vertical Angles/Alternate Interior Angles

Part C) Distance from B to E is 125; Distance from P to E is 275.

Step-by-step explanation:

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Solve the equation x/4 - 7 = y got x

Lina20 [59]

Answer:

x = y · 11

Step-by-step explanation:

Since x is divided by 4, we have to multiply each side by 4.

x-7=y · 4 To get x by itself, add 7 to each side to get x = y · 11

4 0

3 years ago

The level of the tide in a harbor changed from 9 1/4 ft to 4 1/2 ft above sea level over a period of 3 1/4 hr

kumpel [21]

Answer:

The answer is " $\bold{- \frac{19}{13} \ / hour}$ "

Step-by-step explanation:

Please find the complete question in the attached file.

Calculating the difference of $9 \frac{1}{4}\ and \ 4 \frac{1}{2}$ dividing the value by $3 \frac{1}{4}$

$\to 4 \frac{1}{2} - 9 \frac{1}{4}\\\\ \to \frac{9}{2} - \frac{37}{4}\\\\ \to \frac{18-37}{4} \\\\ \to -\frac{19}{4}\\\\\to -\frac{19}{4} \div \frac{13}{4}= -\frac{19}{4} \times \frac{4}{13} = -\frac{19}{13}\\$

3 0

2 years ago

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 is the value of x? (1) x2+x+10=16 (2) x=4y4+2y2+2?

alexandr402 [8]

X equals the square root of 6-x

x= the eqaution for number 2

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3 years ago

How do you add decimals? I need 2 Examples (not from the internet)

zalisa [80]

well These are the Images on how to add decimals.

4 0

2 years ago

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