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2 years ago

14

Janette went to Tire Mart and paid $122.57 per tire for 4 tires and a flat rate fee for installation for a total of $555.28 befo

re taxes. Monique went to Fast Wheels and paid $136.71 per tire for 4 tires and a flat rate fee for installation for a total of $585.34 before taxes. The difference in the amount that they paid in installation charges is

1 answer:

qaws [65]2 years ago

6 0

122.57x4=490.28 is Janette 555.28-490.28=$65 for installation.

Monique 136.71x4=546.84

585.34-546.84=$39 for installation

65-39=$26 is the answer

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Answer:

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Step-by-step explanation:

(a)

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

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<em>X</em> = 610

<em>n</em> = 1000

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

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

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).

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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).

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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).

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

The width of the interval is affected by:

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Standard deviation.

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

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

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Answer:

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So the equation is:

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The ball hits the ground when when $h(t) = 0$

So

$4.5 + 37.62t -16t ^ 2 = 0$

We use the quadratic formula to solve the equation for t

For a quadratic equation of the form

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The quadratic formula is:

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In this case

$a= -16\\\\b=37.62\\\\c=4.5$

Therefore

$t=\frac{-37.62\±\sqrt{(37.62)^2 -4(-16)(4.5)}}{2(-16)}$

$t_1=-0.114\ s\\\\t_2=2.47\ s$

We take the positive solution.

Finally the ball takes 2.47 seconds to touch the ground

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2 years ago