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3 years ago

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A small bottle of lemonade is half the size of a large bottle. The small bottle costs 80p and the large bottle costs £1.50. Amir

buys 4 small bottles. How much would he save if he buys the same amount of lemonade in large bottles?

2 answers:

Leni [432]3 years ago

4 0

80p = 80 cents

£1.50 = $1.50.

(changing the signs, that's it)

So if a small bottle is half the size of a large bottle, and is 80 cents.

Amir buys 4 bottles, so 0.80 * 4 = 320 = $3.20.

Now 4 small bottles is the same as 2 large ones, so if he bought 2 large bottles: $1.50 * 2 = $3.00.

$3.20 - $3.00 = 0.20 cents.

Amir would've saved 0.20

P.S. What is "p" in European money

lapo4ka [179]3 years ago

3 0

the answer to your question is 0.20

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Therefore the rate change of height is $\frac{1}{\pi}$ m/s.

Step-by-step explanation:

Given that a vertical cylinder is leaking water at rate of 4 m³/s.

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Putting the value $\frac{dV}{dt}$

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2 years ago

Read 2 more answers

A research study uses 800 men under the age of 55. Suppose that 30% carry a marker on the male chromosome that indicates an incr

ss7ja [257]

Answer:

a) There is a 12.11% probability that exactly 1 man has the marker.

b) There is a 85.07% probability that more than 1 has the marker.

Step-by-step explanation:

There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so $\pi = 0.30$

(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?

10 men, so $n = 10$

We want to find P(X = 1). So:

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211$

There is a 12.11% probability that exactly 1 man has the marker.

(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?

That is $P(X > 1)$

We have that:

$P(X \leq 1) + P(X > 1) = 1$

$P(X > 1) = 1 - P(X \leq 1)$

We also have that:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282$

So

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493$

Finally

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507$

There is a 85.07% probability that more than 1 has the marker.

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3 years ago