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KIM [24]
2 years ago
8

I will give lots of points please help me

Mathematics
2 answers:
k0ka [10]2 years ago
5 0

Answer:

Part A: 6

Part B: 6√3

Step-by-step explanation:

Part A:

Since one angle measures 30°, and another angle is a right angle measuring 90°, the third angle measures 60°.

In a 30-60-90 the short leg is half of the hypotenuse.

YX = 12/2 = 6

Part B:

Method 1:

In a 30-60-90 triangle, the length of the long leg is √3 times the length of the short leg.

YZ = 6√3

Method 2:

Use the Pythagorean theorem.

a² + b² = c²

6² + (YZ)² = 12²

(YZ)² = 108

YZ = √(108)

YZ = √(36 × 3)

YZ = 6√3

strojnjashka [21]2 years ago
4 0
  • sin30=XY/XZ
  • 1/2=XY/12
  • XY=6

#B

  • Cos30=YZ/XZ
  • √3/2=YZ/12
  • YZ=6√3

Another way is using Pythagorean theorem

  • YZ²=XZ²-XY²
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Suppose that in a random selection of 100 colored​ candies, 28​% of them are blue. The candy company claims that the percentage
Zigmanuir [339]

Answer:

We conclude that the percentage of blue candies is equal to 29​%.

Step-by-step explanation:

We are given that in a random selection of 100 colored​ candies, 28​% of them are blue. The candy company claims that the percentage of blue candies is equal to 29​%.

Let p = <u><em>population percentage of blue candies</em></u>

So, Null Hypothesis, H_0 : p = 29%     {means that the percentage of blue candies is equal to 29​%}

Alternate Hypothesis, H_A : p \neq 29%     {means that the percentage of blue candies is different from 29​%}

The test statistics that will be used here is <u>One-sample z-test for</u> <u>proportions</u>;

                         T.S.  =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of blue coloured candies = 28%

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So, <u><em>the test statistics</em></u> =  \frac{0.28-0.29}{\sqrt{\frac{0.29(1-0.29)}{100} } }

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Now, at a 0.10 level of significance, the z table gives a critical value of -1.645 and 1.645 for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of z, <u><em>so we insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the percentage of blue candies is equal to 29​%.

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