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Hitman42 [59]
3 years ago
15

2. The following data set shows the number of points scored by rugby team in season of matches: (5

Mathematics
1 answer:
shutvik [7]3 years ago
3 0

Answer:

<u><em>Hope it helped :)</em></u>

Step-by-step explanation:

a)Mean  = \frac{sum \ of\ the\ data} {number\ of\ data} = \frac{598}{21} = 28.48\\\\b) Median = middle\ number = 28\\\\c)Mode = most \ frequently \ occured \ number = 25\\\\d) Range = 48 - 5 = 43\\\\e)Inter Quartile Range = Q_3 - Q_1 = 16

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I really need to know how long it takes the large pipe on its own
Andreyy89

s = hours it takes the small pipe on its own.


L = hours it takes the large pipe on its own.


we know the small pipe can do it in 11 hours total, so s = 11, so, how much of the whole job has the small done in 1 hour only?


well, if it takes 11 hours to do the whole thing, in 1 hour it has done only 1/11 of the whole job.


we know both pipes working together can do it in 6 hours flat. So in 1 hour only, both of them have done only 1/6 of the whole thing.


in that 1 hour, how much has the large one done? well, it has only done 1/L of the job.


\bf \stackrel{\textit{how much has it been done by both in 1 hour}}{\stackrel{\stackrel{small}{rate}}{\cfrac{1}{s}}+\stackrel{\stackrel{large}{rate}}{\cfrac{1}{L}}~~=~~\stackrel{job}{\cfrac{1}{6}}} \\\\\\ \cfrac{1}{11}+\cfrac{1}{L}=\cfrac{1}{6}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{66L}}{6L+66=11L}\implies 66=5L \\\\\\ \cfrac{66}{5}=L\implies 13\frac{1}{5}=L\impliedby \textit{13 hours and 12 minutes}


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7 0
4 years ago
An AISI 1040 cold-drawn steel tube has an OD 5 50 mm and wall thickness 6 mm. What maximum external pressure can this tube withs
lukranit [14]

Answer:

82.79MPa

Step-by-step explanation:

Minimum yield strength for AISI 1040 cold drawn steel as obtained from literature, Sᵧ = 490 MPa

Given, outer radius, r₀ = 25mm = 0.025m, thickness = 6mm = 0.006m, internal radius, rᵢ = 19mm = 0.019m,

Largest allowable stress = 0.8(-490) = -392 MPa (minus sign because of compressive nature of the stress)

The tangential stress, σₜ = - ((r₀²p₀)/(r₀² - rᵢ²))(1 + (rᵢ²/r²))

But the maximum tangential stress will occur on the internal diameter of the tube, where r = rᵢ

σₜₘₐₓ = -2(r₀²p₀)/(r₀² - rᵢ²)

p₀ = - σₜₘₐₓ(r₀² - rᵢ²)/2(r₀²) = -392(0.025² - 0.019²)/2(0.025²) = 82.79 MPa.

Hope this helps!!

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