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scZoUnD [109]
3 years ago
10

Assume that BC is tangent to circle A. Find the value of x. Round your answer to the nearest tenth

Mathematics
1 answer:
Natalija [7]3 years ago
3 0

Answer:

x ≈ 9.6

Step-by-step explanation:

Δ ABC is right with ∠ C = 90° ( angle between tangent and radius )

Using Pythagoras' identity in the right triangle

x² + 14² = 17²

x² + 196 = 289 ( subtract 196 from both sides )

x² = 93 ( take the square root of both sides )

x = \sqrt{93} ≈ 9.6 ( to the nearest tenth )

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3 years ago
Nvm don't answer this
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2 years ago
When a distribution is mound-shaped symmetrical, what is the general relationship among the values of the mean, median, and mode
yuradex [85]

Answer:

The mean, median, and mode are approximately equal.

Step-by-step explanation:

The mean, median, and mode are <em>central tendency measures</em> in a distribution. That is, they are measures that correspond to a value that represents, roughly speaking, "the center" of the data distribution.

In the case of a <em>normal distribution</em>, these measures are located at the same point (i.e., mean = median = mode) and the values for this type of distribution are symmetrically distributed above and below the mean (mean = median = mode).

When a <em>distribution is not symmetrical</em>, we say it is <em>skewed</em>. The skewness is a measure of the <em>asymmetry</em> of the distribution. In this case, <em>the mean, median and mode are not the same</em>, and we have different possibilities as the mentioned in the question: the mean is less than the median and the mode (<em>negative skew</em>), or greater than them (<em>positive skew</em>), or approximately equal than the median but much greater than the mode (a variation of a <em>positive skew</em> case).  

In the case of the normal distribution, the skewness is 0 (zero).

Therefore, in the case of a <em>mound-shaped symmetrical distribution</em>, it resembles the <em>normal distribution</em> and, as a result, it has similar characteristics for the mean, the median, and the mode, that is, <em>they are all approximately equal</em>. So, <em>the </em><em>general</em><em> relationship among the values for these central tendency measures is that they are all approximately equal for mound-shaped symmetrical distributions, </em>considering they have similar characteristics of the <em>normal distribution</em>, which is also a mound-shaped symmetrical distribution (as well as the t-student distribution).

5 0
3 years ago
How to divide 457 by 4
Basile [38]
Do you need to show your work? the answer is 114.25
3 0
3 years ago
Read 2 more answers
Can someone help answer this
kolezko [41]

Answer:

v = 55

Step-by-step explanation:

Plug in the values provided into the equation

v = u + at

v = 23 + 8 * 4

Solve

v = 23 + 8 * 4

v = 23 + 32

<u>v = 55</u>

6 0
3 years ago
Read 2 more answers
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