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yaroslaw [1]
3 years ago
14

A unit of electricity costs 13.2 pence. On average, tanya uses 90 units of electricity per week. She pays for her bill in 12 wee

ks. How much will her electricity bill be then?
Mathematics
2 answers:
drek231 [11]3 years ago
8 0

Answer:

14256 pence

Step-by-step explanation:

Cost of electricity of a bill for 1 unit= 13.2 pence

Number of units used by Tanya in one week= 90 units

Total number of units in 12 weeks= 90×12= 1080 units

Cost of 1080 units= 1080×13.2= 14256 pence

Hence the total cost of a bill= 14256 pence

Hence, the correct answer is 14256 pence.

Scrat [10]3 years ago
7 0

Her electricity bill will be of 14256 pence.

<em><u>Explanation</u></em>

On average, Tanya uses 90 units of electricity per week.

So, the <u>total units of electricity in 12 weeks</u> will be: (90*12)=1080 units

One unit of electricity costs 13.2 pence.

Thus, <u>the total cost for 1080 units</u> will be:  (13.2*1080)pence= 14256 pence.

So, her electricity bill will be of 14256 pence.

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1 year 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
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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})

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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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