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vovangra [49]
2 years ago
15

Plz help find each circle measure !!!!!

Mathematics
2 answers:
Y_Kistochka [10]2 years ago
7 0

9514 1404 393

Answer:

  ED = 26; CF = 13; arc ED = 136°; arc HD = 68°; arc CE = 88°

  QU = 8; QR= 16; arc ST = 114°; arc QR = 114°; arc XT = 57°

Step-by-step explanation:

Chords the same distance from the center of the circle are the same length. Chords of the same length intercept arcs of the same angle measure. A radius perpendicular to a chord bisects it and the arc it subtends. The sum of arcs in a circle is 360°. These facts are used to solve the problems here.

__

<u>Circle J</u>

ED = 2×GD = 26

CF = GD = 13

arc ED = arc CD = 136°

arc HD = 1/2 arc CD = 68°

arc CE = 360° - 2×136° = 88°

__

<u>Circle Y</u>

QU = 1/2×ST = 8

QR = ST = 16

arc ST = arc QR = 1/2(360° -34° -98°) = 114°

arc XT = 1/2 arc ST = 57°

Masja [62]2 years ago
6 0

Answer:

for the first one it is

ED= 26

CF= 13

mED= 136

mHD= 68

mCE= dont know this one but the other ones im positive their right

Step-by-step explanation:

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Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

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Step-by-step explanation:

Let <em>X</em> = number of seconds in the batch.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.31.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

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(a)

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P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

              =1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776

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(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

                            ={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

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Justin is redoing his bathroom floor with tiles measuring 5in. By 14 in. The floor has an area of 8,500 square inches. What is t
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I am joyous to assist you anytime.

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