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Mrrafil [7]
3 years ago
14

Old Grandpa Joe left half his money to his granddaughter and half that amount to his grandson.

Mathematics
1 answer:
Usimov [2.4K]3 years ago
7 0

Answer:

£12000

Step-by-step explanation:

Granddaughter -- 1/2 of x

Grandson -- 1/2(1/2 of x)

Brother -- 1/6 of x

Dogs' home -- £1000

Therefore,

\frac{x}{2} + \frac{x}{4} + \frac{x}{6} + 1000 = x

x = 12000

Thenks and mark me brainliest :)

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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
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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