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svlad2 [7]
3 years ago
8

54) |1-2/3x|<1

Mathematics
1 answer:
Naily [24]3 years ago
4 0

|a| < b\iff a < b\ \wedge\ a > -b\\\\|a| > b\iff a > b\ \vee\ a < -b

\left|1-\dfrac{2}{3}x\right| < 1\iff1-\dfrac{2}{3}x < 1\ \wedge\ 1-\dfrac{2}{3}x > -1\ \ \ |\text{subtract 1 from both sides}\\\\-\dfrac{2}{3}x < 0\ \vedge\ -\dfrac{2}{3}x > -2\ \ \ \ |\text{change the signs}\\\\\dfrac{2}{3}x > 0\ \wedge\ \dfrac{2}{3}x < 2\ \ \ \ |\text{multiply both sides by 3}\\\\2x > 0\ \wedge\ 2x < 6\ \ \ \ |\text{divide both sides by 2}\\\\x > 0\ \wedge\ x < 3\to0 < x < 3\to x\in(0,\ 3)

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What is the measure of each angle of a regular 24-gon? If necessary, round to the<br> nearest tenth.
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4 0
3 years ago
Someone claims that the breaking strength of their climbing rope is 2,000 psi, with a standard deviation of 10 psi. We think the
denpristay [2]

Answer:

The sample size must be greater than 37 if we want to reject the null hypothesis.

Step-by-step explanation:

We are given that someone claims that the breaking strength of their climbing rope is 2,000 psi, with a standard deviation of 10 psi.

Also, we are given a level of significance of 5%.

Let \mu = <u><em>mean breaking strength of their climbing rope</em></u>

SO, Null Hypothesis, H_0 : \mu = 2,000 psi       {means that the mean breaking strength of their climbing rope is 2,000 psi}

Alternate Hypothesis, H_A : \mu < 2,000 psi      {means that the mean breaking strength of their climbing rope is lower than 2,000 psi}

Now, the test statistics that we will use here is One-sample z-test statistics as we know about population standard deviation;

                         T.S.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = ample mean strength = 1,997.2956 psi

            \sigma = population standard devaition = 10 psi

            n = sample size

Now, at the 5% level of significance, the z table gives a critical value of -1.645 for the left-tailed test.

So, to reject our null hypothesis our test statistics must be less than -1.645 as only then we have sufficient evidence to reject our null hypothesis.

SO,  T.S. < -1.645   {then reject null hypothesis}

         \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < -1.645

         \frac{1,997.2956-2,000}{\frac{10}{\sqrt{n} } } < -1.645

         (\frac{1,997.2956-2,000}{10}) \times {\sqrt{n} } } < -1.645

          -0.27044 \times \sqrt{n}< -1.645

               \sqrt{n}> \frac{-1.645}{-0.27044}

                 \sqrt{n}>6.083

                  n > 36.99 ≈ 37.

SO, the sample size must be greater than 37 if we want to reject the null hypothesis.

7 0
3 years ago
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