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KengaRu [80]
3 years ago
9

Anya makes a scale drawing of a bug. The scale of the drawing’s length to the actual bug’s length is 2 inches: 1/8 inch. The act

ual bug is 3/4 inch long. How long is Anya’s drawing of the bug.
Mathematics
1 answer:
ioda3 years ago
7 0
<h2>Hello!</h2>

The answer is: The Anya's drawing is 6 inches long.

Have a nice day!

<h2>Why?</h2>

To solve this problem, we need to consider the drawing's length with respect to the actual bug's size, so, the scale factor is:

2in:\frac{1}{8}in\\\\2x(in):\frac{3}{4}in

So, how long is Anya's drawing of the bug?

We can find it, using the following equation:

(2x)in=\frac{2in*\frac{3}{4}in}{\frac{1}{8}in}=\frac{\frac{6}{4}in}{\frac{1}{8}in}=\frac{\frac{3}{2}in }{\frac{1}{8}in}\\\\(2x)in=\frac{\frac{3}{2}in }{\frac{1}{8}in }=\frac{3}{2}in*\frac{8}{1}in=12in^{2}\\\\(2x)in=12in^{2} \\\\x=\frac{12in^{2} }{2in}=6in

Hence, the Anya's drawing is 6 inches long.

Have a nice day!

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The objective is to test if the two machines are filling the bottles with a net volume of 16.0 ounces or if at least one of them is different.

The parameter of interest is μ₁ - μ₂, if both machines are filling the same net volume this difference will be zero, if not it will be different.

You have one sample of ten bottles filled by each machine and the net volume of the bottles are measured, determining two variables of interest:

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Sample 2 - Machine 2

X₂: Net volume of a plastic bottle filled by the machine 2

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With p-values 0.4209 (for X₁) and 0.6174 (for X₂) for the normality test and using α: 0.05, we can say that both variables of interest have a normal distribution:

X₁~N(μ₁;δ₁²)

X₂~N(μ₂;δ₂²)

Both population variances are unknown you have to conduct a homogeneity of variances test to see if they are equal (if they are you can conduct a pooled t-test) or different (if the variances are different you have to use the Welche's t-test)

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Using a statistic software I've calculated the test

F_{H_0}= 1.41

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The p-value is greater than the significance level, so the decision is to not reject the null hypothesis then you can conclude that both population variances for the net volume filled in the plastic bottles by machines 1 and 2 are equal.

To study the difference between the population means you can use

t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~~t_{n_1+n_2-2}

The hypotheses of interest are:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

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The p-value for this test is: 0.882433

The p-value is greater than the significance level, the decision is to not reject the null hypothesis.

Using a significance level of 5%, there is no significant evidence to reject the null hypothesis. You can conclude that the population average of the net volume of the plastic bottles filled by machine one and by machine 2 are equal.

I hope this helps!

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