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solmaris [256]
3 years ago
15

Five friends equally shared half of one large pizza and 1/4 of another large pizza. what fraction of each pizza did each friend

get? how do the two amounts compare each other?
Mathematics
1 answer:
attashe74 [19]3 years ago
3 0
1/2 divide by 5= 1/10
1/4 divide by 5=  1/20                              so    1/10 is also 2/20+1/20=3/20

so the answer is 3/20 of a pizza
 hope this helped and plz leave a thank you

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Answer:

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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:

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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.

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

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

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

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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.

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

  • The data should be continuous.
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