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Vlada [557]
3 years ago
13

c%7B4%7D%7B5%7D%20%5Ctimes%20%20%5Cfrac%7B3%7D%7B4%7D%20%20" id="TexFormula1" title=" \frac{2}{3 } \times \frac{5}{7} \times \frac{4}{5} \times \frac{3}{4} " alt=" \frac{2}{3 } \times \frac{5}{7} \times \frac{4}{5} \times \frac{3}{4} " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
VikaD [51]3 years ago
5 0

Answer:

\frac{2}{3}*\frac{5}{7}*\frac{4}{5}*\frac{3}{4}\\=\frac{2}{7}

Step-by-step explanation:

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gtnhenbr [62]

Answer:

c. 6

Step-by-step explanation:

7 0
2 years ago
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A cylindrical 2014-t6 aluminum alloy bar is subjected to compression-tension stress cycling along its axis; results of these tes
Pani-rosa [81]

Complete part of Question:

Assume a factor of safety of 3.0

The Maximum stress versus logarithm of the number of cycles to fatigue failure curve for 2014-T6 Aluminium bar is attached to this solution

Answer:

The maximum allowable load amplitude is 6403.33 N

Step-by-step explanation:

diameter of the bar, d = 12 mm

d = 0.012 m

Cross sectional area of the bar, A = πd²/4

A = π*0.012²/4

A = 0.000113 m²

From the Maximum stress versus logarithm of the number of cycles to fatigue failure for 2014-T6 Aluminium bar attached to this solution;

At 10⁷ cycles, Maximum stress, S = 170 MPa

S = 170 * 10⁶ Pa

Factor of safety, N = 3.0

The tensile stress is given by the formula:

\sigma = S/N\\\sigma = \frac{170 * 10^{6} }{3} \\\sigma = 56.67 * 10^{6} N/m^{2}

The tensile stress is also given by:

\sigma = F/A\\\sigma = F/0.000113

F/0.000113 = 56.67 * 10⁶

F = 56.67 * 10⁶ * 0.000113

F = 6403.33 N

The maximum allowable load amplitude is 6403.33 N

3 0
3 years ago
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Can I have more help? Thank you xtra points!
Feliz [49]

Answer:

135

Step-by-step explanation:

8 0
3 years ago
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I will vote u brainliest if u answer so pls do
Lera25 [3.4K]

Answer: For this, you have to calculate the Unit value,

Unit value of 1rst one,  4.99/6 = 0.83

Unit value of 2nd one, 7.99/10 = 0.79

As, 2nd one is cheaper, so it is better :))

Step-by-step explanation:

8 0
3 years ago
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A national survey of companies included a question that asked whether the company had at least one bilingual telephone operator.
nirvana33 [79]

Answer:

The first option is correct. Option A is correct.

LCL = 0.270, and UCL = 0.397

80% Confidence interval = (0.270, 0.397)

Step-by-step explanation:

The data for Y and N for the 90 companies is attached to this solution provided.

Y represents companies with at least 1 bilingual operator and N represents companies with no bilingual operator.

The number of Y in the data = 30

Hence, sample proportion of companies with at least one bilingual operator = (30/90) = 0.3333

Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion = 0.3333

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 80% confidence level for sample size of 90 is obtained from the z-tables.

Critical value = 1.280

Standard error of the mean = σₓ = √[p(1-p)/n]

p = sample proportion

n = sample size = 90

σₓ = √[0.3333×0.6667)/90] = 0.0496891568 = 0.04969

80% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.3333 ± (1.28 × 0.04969)

CI = 0.3333 ± 0.0636021207

80% CI = (0.2696978793, 0.3969021207)

80% Confidence interval = (0.270, 0.397)

Hope this Helps!!!

3 0
3 years ago
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