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Norma-Jean [14]
2 years ago
9

Sean got 4 over 5 pound of honey from a hive. He calculated the quotient of 4 over 5 ÷ 3 over 8. Which statement shows the quoti

ent and gives the most likely explanation of what Sean was trying to determine?
He can make 2 and 2 over 15 bottles of honey with 3 over 8 pound of honey in each bottle.
He can make 12 over 40 of a bottle of honey with 3 over 8 pound of honey in each bottle.
He can get 2 and 2 over 15 pounds of honey if he divides the honey into 3 over 8-pound portions.
He can get 12 over 40 pound of honey if he divides the honey into 3 over 8-pound portions.
Mathematics
1 answer:
Alexandra [31]2 years ago
8 0
The answer is He can make 12/40 of a bottle of honey with 3 over 8 pounds of honey.The reason is because it kinda look more sense to compare in the question to the 2nd answer
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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.

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2 years ago
PLEASE HELP ASAP 30 POINTS ):
ladessa [460]

Answer:

c (0,-4)

Step-by-step explanation:

-5x+y = -4

4x - 4y =16

Solve the first equation for y since we are using substitution.

-5x+y = -4

Add 5x to each side

-5x+5x+y = -4+5x

y = 5x-4

Substitute this equation y = 5x-4 into the second equation.

4x -4(5x-4) = 16

Distribute the -4

4x - 4(5x) -4(-4) = 16

4x-20x +16 = 16

Combine like terms

-16x +16 =16

Subtract 16 from each side

-16x+16-16 = 16-16

-16x =0

Divide by -16

x=0


But we still need to find y

y = 5x-4

y = 5(0) -4

y = -4

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Look at the figure. is the intersection of plane _____ and plane QRST.
ivanzaharov [21]
Plane UVWX and plane QRST
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2 years ago
In 2000, U.S. Airways burned about 115 litres of fuel per passenger. In 2014 that dropped
ankoles [38]

Answer:

22% reduction

Step-by-step explanation:

Find the percent decrease by dividing the difference in amounts by the original amount.

Find the difference:

115 - 90 = 25

Divide this by the original amount:

25/115

= 0.217

So, the percent decrease is 21.7%. Round this to the nearest whole percentage:

= 22

The percent reduction in fuel used was approx. 22%

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3 years ago
Each of the students im Romi's class raised at least $25 during the jump-a-thon. What is the least amount of money the class rai
EastWind [94]
If each of the 28 students made at least $25, you would multiply 28 and 25 together to obtain the least amount of money the class raised. That gets, 28x25 = 700. The class made at least $700.
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3 years ago
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