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1 year ago

Mr. Donley drove 504 km in each direction on a vacation

Mathematics

1 answer:

mestny [16]1 year ago

6 0

Answer:

Data that we know:

He drove 504km in each direction.

Going, the average speed was 14km/h more than returning.

The total trip lasted 21 hours.

Ok, to solve this first:

Let's define the variables:

Sg = Average speed when going.

Sr = Average speed when returning.

Then we can write one of our relationships as:

Sg = Sr + 14km/h.

Now, you can recall the relation:

Time = Distance/speed.

Then we can write the equation that represents the total time of the travel as:

504km/Sg + 504km/Sr = 21h

Now we can replace the Sg by Sr + 14km/h, and get:

504km/(Sr + 14km/h) + 504km/Sr = 21h.

Now we must multiplacate by each denominator, in order to remove them:

504km*Sr + 504km*(Sr + 14km/h) = 21h*Sr*(Sr + 14km/h)

Sr*1008km + 7056km^2/h = 21h*Sr^2 + 294km*Sr

Now we can write a cuadratic equation, where i will ignore the units so it is easier to read, as:

21*Sr^2 -714*Sr - 7056 = 0

The solutions are given by the Bhaskara formula:

$Sr = \frac{714 +- \sqrt{714^2 - 4*21*(-7056)} }{2*21} = \frac{714 +- 1050}{42}$

We have two solutions, one negative and one positive, and because Sr represents an average speed, it must be a positive number, then we choose the positive solution:

Sr = (714 + 1050)/42 km/h = 42km/h

Now we have the average speed for the returning, with this we can find Sg.

Sg = Sr + 14km/h = 42km/h + 14km/h = 56km/h.

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A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the

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Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).

Step-by-step explanation:

(a)

Let X = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

X = 610

n = 1000

Confidence level = 90%

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{610}{1000}=0.61$

The critical value of z for a 90% confidence level is:

$z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645$

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)$

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

Interpretation:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let X = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

X = 440

n = 1000

Confidence level = 95%

The (1 - α)% confidence interval for population proportion is:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

The sample proportion is:

$\hat p=\frac{X}{n}=\frac{440}{1000}=0.44$

The critical value of z for a 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

$CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)$

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend is (0.42, 0.46).

Interpretation:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

The confidence level
Sample size
Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of z in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).