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3 years ago

Help with trigonometry

Mathematics

1 answer:

diamong [38]3 years ago

4 0

Answer:

cos ∠CBD = - 4√41 / 41

Step-by-step explanation:

AB = 4 ΔABC = (4 x AC) / 2 = 10

AC = 5

BC = √5² + 4² = √41

cos ∠CBD = cos (180° - ∠CBA) = cos 180° cos ∠CBA + sin 180° sin ∠CBA

(cos 180° = - 1 sin 180° = 0 cos ∠CBA = 4 / √41 )

cos ∠CBD = (-1) x cos ∠CBA = - 4 / √41 = - 4√41 / 41

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STALIN [3.7K]

Answer:

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3 years ago

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bulgar [2K]

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Step-by-step explanation:

use PEMDAS to solve

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3 years ago

Nancy plans to deposit $1,000 quarterly for 35 years at 8.25% Interest, compounded monthly. How much will she

melomori [17]

Answer:

$17,771.92

Step-by-step explanation:

A = P (1 + r/n)^nt

A = Future value (the answer you're trying to find)

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T = Time (35)

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3 years ago

Two sides of a rectangle iPhones are five and 5/8 feet long other two sides of 6 1/4 feet long what is the perimeter

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Step-by-step explanation:

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4 years ago

The Ball Corporation's beverage can manufacturing plant in Fort Atkinson, Wisconsin, uses a metal supplier that provides metal w

vampirchik [111]

Answer:

The 95 percent confidence interval for the true mean metal thickness is between 0.2903 mm and 0.2907 mm

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{0.000586}{\sqrt{59}} = 0.0002$

The lower end of the interval is the sample mean subtracted by M. So it is 0.2905 - 0.0002 = 0.2903 mm

The upper end of the interval is the sample mean added to M. So it is 0.2905 + 0.0002 = 0.2907 mm

The 95 percent confidence interval for the true mean metal thickness is between 0.2903 mm and 0.2907 mm

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3 years ago

Help with trigonometry​

Help with trigonometry