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2 years ago

Find the angle between the given vectors to the nearest tenth of a degree.

Mathematics

2 answers:

andreyandreev [35.5K]2 years ago

8 0

Answer:

20.3

Step-by-step explanation:

Pavel [41]2 years ago

8 0

Answer:

Step-by-step explanation:

u=6i-j

v=7i-4j

u.v=|6i-j||7i-4j| cos α,where α is the angle between u & v.

(6i-j).(7i-4j)=√(6²+(-1)²)√(7²+(-4)²) cos α

(6)(7)+(-1)(-4)=√37√65 cos α

42+4=√37√65 cos α

$\cos \alpha =\frac{46}{\sqrt{37} \sqrt{65} } \\\alpha =\cos^{-1}\frac{46}{\sqrt{37} \sqrt{65} } \approx 20.28^\circ \approx 20.3^\circ$

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UkoKoshka [18]

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a) The minimum sample size is 601.

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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.

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This is n for which M = 0.04. So

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

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

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b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

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

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$\sqrt{n} = \frac{1.96*0.5}{0.02}$

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$n = 2401$

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2 years ago

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Sidana [21]

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