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3 years ago

What is the ratio of 40cm and 18mm

Mathematics

1 answer:

snow_tiger [21]3 years ago

5 0

40 cm = 400 mm, so 400 mm / 18 mm = 200:9, or 22.22:1.

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90 is the largest integer

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3 years ago

What cryptographic method, first proposed in the mid-1980s, makes use of sloping curves instead of large prime numbers?

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3 years ago

Complete the steps to find the value of x

vampirchik [111]

72, 96, 84, 12

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3 years ago

a truck weighs 5,400 pounds. An open-wheel race car weighs 1/4 as much. how much does the race car weigh

gtnhenbr [62]

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3 0

3 years ago

Read 2 more answers

A quality-conscious disk manufacturer wishes to know the fraction of disks his company makes which are defective. Step 2 of 2 :

const2013 [10]

Answer:

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Suppose a sample of 1067 floppy disks is drawn. Of these disks, 74 were defective.

This means that $n = 1067, \pi = \frac{74}{1067} = 0.069$

80% confidence level

So $\alpha = 0.2$ , z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$ , so $Z = 1.28$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 - 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.059$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.069 + 1.28\sqrt{\frac{0.069*0.931}{1067}} = 0.079$

The 80% confidence interval for the population proportion of disks which are defective is (0.059, 0.079).

7 0

2 years ago