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3 years ago

13

What is the answer to iii) ??

1 answer:

Alexxandr [17]3 years ago

8 0

Since there is an expression in the square root, it must be non-negative. So:

$f(x)=\sqrt{12-3x}\\\\ \Longrightarrow 12-3x\geq0\\\\ 3x\leq12\\\\ x\leq\dfrac{12}{3}\\\\ x\leq4$

Then, the interval is:

$\boxed{x\in~]-\infty,4]}$

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An environmentalist wants to find out the fraction of oil tankers that have spills each month. Step 2 of 2 : Suppose a sample of

rewona [7]

Answer:

The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).

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 333 tankers is drawn. Of these ships, 257 did not have spills.

333 - 257 = 76 have spills.

This means that $n = 333, \pi = \frac{76}{333} = 0.228$

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 - 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.199$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.228 + 1.28\sqrt{\frac{0.228*0.772}{333}} = 0.257$

The 80% confidence interval for the population proportion of oil tankers that have spills each month is (0.199, 0.257).

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Whats 18 over 50 simplified

Harlamova29_29 [7]

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