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Tcecarenko [31]
3 years ago
13

An open-topped glass aquarium with a square base is designed to hold 13.513.513, point, 5 cubic feet of water. what is the minim

um exterior surface area of the aquarium?
Mathematics
1 answer:
ExtremeBDS [4]3 years ago
3 0
Suppose the aquarium's shape is a square prism, with each side of the square having length x, and the aquarium's height is y, so we have: 
<span> x^2 * y = 13.5
</span> <span>y = 13.5/(x^2) 
</span><span> The area of each side is xy, and there are four sides, so that totals 4xy 
</span><span> The area of the bottom is x^2 
</span><span> Total surface area: 
</span><span> A = x^2 + 4xy 
</span><span> A = x^2 + 4x(13.5/(x^2))
</span> <span>A = x^2 + 54x^(-1) 
</span><span> dA/dx = 2x - 54x^(-2) 
</span><span> Set the derivative to zero: 
</span><span> 2x - 54x^(-2) = 0 
</span><span> 2x = 54/(x^2) 
</span><span> 2x^3 = 54 
</span><span> x^3 = 54/2 
</span><span> x^3 = 27 
</span><span> x = cbrt(27) = 3, assuming x is a real number 
</span><span> A = 3^2 + 54*3^(-1) 
</span><span> A = 9 + 18
</span><span> A = 27 square feet</span>
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To compare two programs for training industrial workers to perform a skilled job, 20 workers are included in an experiment. Of t
nikklg [1K]

Answer:

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

p_v =P(t_{(18)}

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

Step-by-step explanation:

Data given and notation

We can calculate the sample mean and deviation with these formulas:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X_{1}=19.1 represent the mean for the sample mean for 1

\bar X_{2}=23.3 represent the mean for the sample mean for 2

s_{1}=4.818 represent the sample standard deviation for the sample 1

s_{2}=5.559 represent the sample standard deviation for the sample 2

n_{1}=10 sample size selected 1

n_{2}=10 sample size selected 2

\alpha represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the average time taken when training under method 1 is less than the average time for Method 2, the system of hypothesis would be:

Null hypothesis:\mu_{1} \geq \mu_{2}

Alternative hypothesis:\mu_{1} < \mu_{2}

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

t=\frac{19.1-23.3}{\sqrt{\frac{4.818^2}{10}+\frac{5.559^2}{10}}}}=-1.805  

P-value

The first step is calculate the degrees of freedom, on this case:

df=n_{1}+n_{2}-2=10+10-2=18

Since is a one sided test the p value would be:

p_v =P(t_{(18)}

Conclusion

If we compare the p value and the significance level assumed \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude the the true mean for method 1 is lower than the mean for the method 2 at 5% of significance

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FromTheMoon [43]
0.3, 0.03,0.003
Because 0.3 is equivalent to 3/10
Because 0.03 is equivalent to 3/100
Because 0.003 is equivalent to 3/1000
4 0
3 years ago
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