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Liula [17]
3 years ago
12

Dave Goes into a car and buys 2 cups of coffee and a piece of cake each cup of coffee cost to pound 50 5K cost £3.85 to the 10th

power mode he thinks he'll get more than £1.50 change
Mathematics
1 answer:
goldenfox [79]3 years ago
4 0

Answer:

Dave will get £1.60 change. Hence Dave is right

Step-by-step explanation:

Complete question

Dave goes into a cafe and buys 2 cups of coffee and a piece of cake.

Each

cup of coffee costs £2.75

The cake costs £2.90

Dave pays with a £10 note.

He thinks he will get more than £1.50 in change.

Is Dave correct?

Solution

Cost of one cup of coffee = £2.75

Cost of two cup of coffee = 2* £2.75 = £ 5.5

Cost of cake =  £2.90

Total amount given by Dave =  £10

Amount left with Dave =  £10 -£ 5.5 - £2.90 = £1.60

Dave will get £1.60 change. Hence Dave is right

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Answer:

2.5 half-lives

Step-by-step explanation:

Find t_{h} in N(t)=N_{0}e^{-kt} given N(t) = 2, N_{0} =64,t=\frac{25}{2}

N(t) is the amount after the time t, N_0 `is the initial amount, t_{h} is the half-life

We know that after half-life there will be twice less the initial quantity:

N(t_{h})=\frac{N_{0} }{2}=N_{0}e^{kt_{h} }.

Simplifying gives \frac{1}{2} =e^{-kt_h} or k = -\frac{In(\frac{1}{2}) }{t_{h} }  .

Plugging this into the initial equation, we obtain that N(t)=N_{0}e^\frac{In(\frac{1}{2}) }{t_{h} }or N(t)= N_{0} (\frac{1}{2})^\frac{t}{t_h}.

Finally, just plug in the given values and find the unknown one.

From 2=64(\frac{1}{2})^\frac{\frac{25}{2} }{t_h}, we have that t_h = \frac{25 In (2)}{2 In (32)\\}

So you should get: t_{h} =\frac{25 In (2)}{2 In (32)\\}=2.5.

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3 years ago
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Five less than one third of a number is two more than half of the same number. Find the number.
I am Lyosha [343]
5 less (-5) than 1/3 of number (n/3) is (=) 2 more than (2+) half of the number (n/2)


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times 6 both sides
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minus 2n both sides
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the number is 18
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Answer:

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Step-by-step explanation:

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Lapatulllka [165]

Answer:

6.56% probability that a real emergency situation exists.

Step-by-step explanation:

We have these following probabilities:

A 0.4% probability that a real emergency situation exists.

A 99.6% probability that a real emergency situation does not exist.

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If an emergency situation does not exist, a 2% probability that the alarm sounds.

The problem can be formulated as the following question:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

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Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem:

What is the probability of a real emergency situation existing, given that the alarm has sounded.

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P(A/B) is the probability of the alarm sounding when there is a real emergency situation. So P(A/B) = 0.95.

P(A) is the probability of the alarm sounding. This is 95% of 0.4%(real emergency situation) and 2% of 99.6%(no real emergency situation). So

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Step-by-step explanation:

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