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Masteriza [31]
3 years ago
8

Does anyone knows how to solve this question??

Mathematics
2 answers:
sergij07 [2.7K]3 years ago
8 0

Answer:

24

Step-by-step explanation:

cos(x) is just a trigonometric function. To find it on a right triangle, you take the "adjacent" side, which should be the side next to the angle x, and divide it by the hypotenuse.

In this case, the adjacent side is 12. We don't know the hypotenuse, but we can figure it out using the Pythagorean Theorem: \sqrt{9^{2}+12^{2}} = \sqrt{225} = 15. So, the hypotenuse is 15.

Thus, cos(x) = 12/15 = 4/5. We want 30 * cos(x), so we get: 30 * (4/5) = 24, and that is our answer.

Ilya [14]3 years ago
7 0

Answer:

24

Step-by-step explanation:

From the mnemonic SOHCAHTOA you would know that cos x = adjacent/hypotenuse.

Adjacent side is given as 12, but we must calculate the hypotenuse.

For that we use Pythagoras.

hypotenuse = √(9²+12²) = 15

Now you can calculate

30×cos(x) = 30×12/15 = 24

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A trail is 13.5 miles Lilongwe. There are markers every 0.25 miles. How many markers are there
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The probability that a farmer is in debt is 0.700. The probability that a farmer is in debt and also lives in the Midwest is 0.2
bogdanovich [222]

Answer:

0.4

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Given that:

P(debt) = P(D) = Probability of being in debt = 0.7

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The probability that a randomly selected farmer lives in the Midwest given that he is in debt is?

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5 0
3 years ago
In a set of 25 aluminum castings, four castings are defective (D), and the remaining twenty-one are good (G). A quality control
lianna [129]

Answer:

The sample space for selecting the group to test contains <u>2,300</u> elementary events.

Step-by-step explanation:

There are a total of <em>N</em> = 25 aluminum castings.

Of these 25 aluminum castings, <em>n</em>₁ = 4 castings are defective (D) and <em>n</em>₂ = 21 are good (G).

It is provided that a quality control inspector randomly selects three of the twenty-five castings without replacement to test.

In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.

The formula to compute the combinations of k items from n is given by the formula:

{n\choose k}=\frac{n!}{k!(n-k)!}

Compute the number of samples that are possible as follows:

{25\choose 3}=\frac{25!}{3!\times (25-3)!}

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The sample space for selecting the group to test contains <u>2,300</u> elementary events.

6 0
3 years ago
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