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Finger [1]
3 years ago
12

Solve the following equations. 3 ⋅ 62^x = 648

Mathematics
1 answer:
gayaneshka [121]3 years ago
7 0

Answer:

x=1.56861

Step-by-step explanation:

62^x=648

By using logarithms it becomes easier to solve this question because of the properties they have

From the property of logarithms

loga^x=xloga

Take logarithms on both sides

log62^x=log648

\Rightarrow xlog62=log648

\Rightarrow x=\frac{log648}{log62}

\Rightarrow x=1.56861

x=1.56861

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

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