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3 years ago

Gareth buys 2 oranges he pays with £1 coin and gets 52p change work out the cost of 1 orange

Mathematics

1 answer:

kap26 [50]3 years ago

5 0

The cost of one orange is 76 p.

<h3>What is division?</h3>

Division is a mathematical operation that is used to determine the quotient of a set of numbers. The sign used to denote division is ÷.

<h3 /><h3>What is the cost of 1 orange?</h3>

The first step is to know how many pence makes a pound.

100 pence = 1 pound

The second step is to convert £1 52 p to pence

1 x 100 + 52 = 152p

The third step is to divide, 152 by 2

152/2 = 74p

To learn how to convert units, please check: brainly.com/question/25993533

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An experimenter flips a coin 100 times and gets 59 heads. Find the 98% confidence interval for the probability of flipping a hea

lara [203]

Answer:

The 98% confidence interval for the probability of flipping a head with this coin is (0.4756, 0.7044).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

An experimenter flips a coin 100 times and gets 59 heads.

This means that $n = 100, \pi = \frac{59}{100} = 0.59$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a p-value of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.59 - 2.327\sqrt{\frac{0.59*0.41}{100}} = 0.4756$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.59 + 2.327\sqrt{\frac{0.59*0.41}{100}} = 0.7044$

The 98% confidence interval for the probability of flipping a head with this coin is (0.4756, 0.7044).

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