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3 years ago

15

A construction company built a scale model of a building. The model was built using a scale of 3 inches = 32 feet. If the buildi

ng is expected to be 200 feet tall, how tall will the model be

1 answer:

Vlad1618 [11]3 years ago

7 0

Let x represent the height of the model.

We have been given that a construction company built a scale model of a building. The model was built using a scale of 3 inches = 32 feet. We are asked to find the height of the model, if the building is expected to be 200 feet tall.

We will use proportions to solve our given problem as:

$\frac{\text{Model height}}{\text{Actual height}}=\frac{\text{Model length}}{\text{Actual length}}$

Upon substituting our given values, we will get:

$\frac{x}{200\text{ ft}}=\frac{3\text{ in}}{\text{32 ft}}$

$\frac{x}{200\text{ ft}}\times 200\text{ ft}=\frac{3\text{ in}}{\text{32 ft}}\times 200\text{ ft}$

$x=3\text{ in}\times 6.25$

$x=18.75\text{ in}$

Therefore, the model will be 18.75 inches tall.

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3 years ago

The number of entrees purchased in a single order at a Noodles & Company restaurant has had an historical average of 1.7 ent

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Answer:

$t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744$

$p_v =P(t_{(47)}>2.744)=0.0043$

If we compare the p value and the significance level given $\alpha=0.02$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=2.1$ represent the mean

$s=1.01$ represent the sample standard deviation

$n=48$ sample size

$\mu_o =1.7$ represent the value that we want to test

$\alpha=0.02$ 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 mean is higher than 1.7, the system of hypothesis would be:

Null hypothesis: $\mu \leq 1.7$

Alternative hypothesis: $\mu > 1.7$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

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

$t=\frac{2.1-1.7}{\frac{1.01}{\sqrt{48}}}=2.744$

P-value

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

$df=n-1=48-1=47$

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

$p_v =P(t_{(47)}>2.744)=0.0043$

Conclusion

If we compare the p value and the significance level given $\alpha=0.02$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is higher than 1,7 entrees per order at 2% of signficance.

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3 years ago

Read 2 more answers

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

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$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

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$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

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3 years ago

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Answer:

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Such that each input can be mapped into only one output.

So for example, if we have a relation that maps the input 2 into two different values:

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Then this is not a function.

In the case of the problem, we have a student as the input, and the hair color as the output.

So we will have something like:

f(student) = blond

And if this student decides to change his/her hair color to red?

Then the function becomes:

f(student) = red

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We also could have the case where a given student has two colors (Californian for example)

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2 years ago