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mars1129 [50]
3 years ago
7

A 10 metre length of Wood is marked out into 10 equal lengths. How many metres from one end is the third mark?

Mathematics
1 answer:
Eduardwww [97]3 years ago
7 0

3 metres from one end to the third mark

Solution:

Total length of wood = 10 metre

Wood is marked into 10 equal length.

$\text{Length of each marking}=\frac{\text{Total length}}{\text{Number of marking}}

                                     $=\frac{10}{10}

                                     = 1 metre

Length of each marking = 1 metre

Number of metres from one end to third mark = 1 + 1 + 1 = 3 metre

The image of the marking is attached below.

Hence 3 metres from one end to the third mark.

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<h2>(f  \circ g)(x) =  |x  +  8|  - 8</h2>

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<h3>(f  \circ g)(x) =  |x  +  8|  - 8</h3>

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A new dental bondlng agent. When bonding teeth, orthodontists must maintain a dry field. A new bonding adhesive(called "Smartbon
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Answer:

(a) <em>H</em>₀: <em>μ ≥ 5.70 </em><em>vs. </em><em>Hₐ</em>:<em>μ < 5.70</em>

(b) The rejection region is (<em>t₀.₀₁,₉</em> <em>≤ -2.821</em>).

(c) The value of the test statistic is -4.33.

(d) The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

Step-by-step explanation:

A hypothesis test should be conducted to determine that the if the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

(a)

The hypothesis is:

<em>H</em>₀: The true mean breaking strength of the new bonding adhesive is not less than 5.70 Mpa, i.e. <em>μ ≥ 5.70</em><em>.</em>

<em>Hₐ</em>: The true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa, i.e. <em>μ < 5.70</em><em>.</em>

(b)

The alternate hypothesis indicates that the hypothesis test is left-tailed.

The rejection region for the left tailed test will be towards the lower tail of the t<em>-</em>distribution curve.

The significance level of the test is: <em>α</em> = 0.01.

The critical value is:

t_{\alpha ,(n-1)}=t_{0.01,(10-1)}=t_{0.01,9}

Use the <em>t-</em>table for the critical value.

t_{\alpha ,(n-1)}=t_{0.01,9}=-2.821

Since rejection region is in the lower tail the critical value will be negative.

Thus, the rejection region is (<em>t₀.₀₁,₉</em> <em>≤ -2.821</em>).

(c)

The test statistic value is:

t=\frac{\bar x-\mu}{s/\sqrt{n}}

Given:

\bar x=5.07\\s=0.46\\n=10\\\mu=5.70

Compute the value of the <em>t</em>-statistic as follows:

t=\frac{\bar x-\mu}{s/\sqrt{n}}=\frac{5.07-5.70}{0.46/\sqrt{10}} =-4.33

The value of the test statistic is -4.33.

(d)

The value of the test is less than the critical value.

t=-4.33

This implies that the test statistic lies in the rejection region.

Hence the null hypothesis will be rejected at 1% significance level.

<u>Conclusion:</u>

As the null hypothesis is rejected it can be concluded that the true mean breaking strength of the new bonding adhesive is less than 5.70 Mpa.

(e)

The conditions required for the <em>t-</em>test for single mean to be valid is:

  • The data should be continuous.
  • The parent population should be normally distributed.
  • The sample should be randomly selected.

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Answer: Option C

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4 0
3 years ago
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