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erica [24]
3 years ago
8

Cos G = 5/12 12/13 5/13 12/5

Mathematics
2 answers:
Gnoma [55]3 years ago
8 0
The answer would be 5/13
Lelu [443]3 years ago
3 0

Answer:

\boxed {\boxed {\sf cos(G)=\frac{5}{13} }}

Step-by-step explanation:

First, recall the trigonometric ratios.

  • sin(θ)=opposite/hypotenuse
  • cos(θ)= adajcent/hypotenuse
  • tan(θ)=opposite/adjacent

The question asks us to find the cosine of G. Therefore, we need the adjacent side and the hypotenuse.

  • Adjacent: 5 is the side next to angle G (12 is opposite, but we don't need that for cosine).
  • Hypotenuse: 13 is the hypotenuse because it is the largest side and opposite the right angle.

Substitute the values into the ratio.

cos(G)= \frac{adjacent}{hypotenuse}

cos(G)=\frac{5}{13}

This fraction cannot be reduced further, so it is the answer.

The cosine of G is <u>5/13</u>

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In this case this "of" means we will multiply the two fractions "4/5 × 3/4" exists 3/5.

<h3>What are fractions?</h3>

That operation exists utilized to 'simplify' a fraction (reduce it to lower terms).

When there exists no number that goes into both the top and bottom numbers, then the fraction is in simplest form (lowest terms).

We are looking for 4/5 of 3/4. In this case this "of" means we will multiply the two fractions. To multiply fractions, we multiply

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4/5 × 3/4 exists 4 × 3 on top and 5 × 4 on the bottom.

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A Packaging Company produces boxes out of cardboard and has a specified weight of 16 oz. A random sample of 36 boxes yielded a s
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Answer:

The margin of error is of 0.73 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.

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 - 0.99}{2} = 0.005

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

That is z with a pvalue of 1 - 0.005 = 0.995, so Z = 2.575.

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 = 2.575\frac{1.7}{\sqrt{36}} = 0.73

The margin of error is of 0.73 oz.

The lower end of the interval is the sample mean subtracted by M. So it is 15.3 - 0.73 = 14.57 oz.

The upper end of the interval is the sample mean added to M. So it is 15.3 + 0.73 = 16.03 oz.

The 99% confidence interval for the true mean weight of the boxes is between 14.57 oz and 16.03 oz. This means that we are 99% sure that the true mean weight of all boxed produced by the Packaging Company is between these two values, and that the specified weight is in this interval.

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3 years ago
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