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Maslowich
3 years ago
10

W = t(r1 + r2) for t

Mathematics
1 answer:
Firlakuza [10]3 years ago
7 0

Answer:

t = w/(r1 + r2)

Step-by-step explanation:

ngl, I guessed on edgenuity and it ended up being the right answer lol.

Just remember to isolate the variable and work your way through different ideas to get the right answer. There are two methods to get the right answer, but I'll use the division method (theres also the distributive property but its more work).

start--> w=t(r1+r2)

divide both sides by (r1 +r2)

answer: w/(r1 + r2) = t

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 Alternative Hypothesis : H₁ : μ < 5.5

b) The test statistic

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c) P - value lies between in these intervals

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Step-by-step explanation:

<u><em>Step( i )</em></u>:-

Given data the Population mean 'μ' = 5.5

The small sample size 'n' = 16

The sample mean (x⁻) = 5.25

Given the  percentage of SiO2 in a sample is normally distributed with a sigma of 0.3.

<u><em> Null hypothesis : H₀ : μ = 5.5</em></u>

<u><em>  Alternative Hypothesis : H₁ : μ < 5.5</em></u>

 Level of significance ∝ = 0.01

<u><em>Step(ii)</em></u>:-

 The test statistic

                              t = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }

                             t = \frac{5.25 -5.5}{\frac{0.3}{\sqrt{16} } }

On calculation , we get

                            t = -3.33

                           |t| = |-3.33| = 3.33

<u><em>Step(iii)</em></u>:-

<u><em>P - value</em></u>

<u><em>The degrees of freedom γ = n-1 = 16-1 =15</em></u>

The calculated value t = 3.33 (check t-table) lies between the 0.001 to 0.005

0.001 < P < 0.005

<u>Condition(i)</u>

P - value < ∝ then reject H₀

<u>Condition(ii)</u>

P - value > ∝ then Accept H₀

we observe that  0.001 < P < 0.005

P- value < 0.01

we rejected  H₀

<em>(or)</em>

The tabulated value  = 2.60 at 0.01 level of significance with '15' degrees of freedom

The calculated value t = 3.33 > 2.60 at 0.01 level of significance with '15' degrees of freedom

The null hypothesis is rejected

<u><em>Conclusion</em></u>:-

Accepted Alternative hypothesis H₁

The Claim that the true average is smaller than 5.5

<u><em></em></u>

             

4 0
3 years ago
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