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3 years ago

5

A plane flies at a steady rate of 40 mph while a second plane flies at a steady rate of 100 mph. The planes are 200 miles apart

and fly directly toward each other. In how many hours will they meet?

2 answers:

Brums [2.3K]3 years ago

7 0

Answer:

$1\ \text{hour}\ 25\ \text{minutes and }12\ \text{seconds}$

Step-by-step explanation:

The combined distance that both planes will cover is 200 miles

Time taken by both planes will be the same (t)

$\text{Distance}=\text{Speed}\times \text{Time}$

So, the combined distance is

$40t+100t=200\\\Rightarrow 140t=200\\\Rightarrow t=\dfrac{200}{140}\\\Rightarrow t=1.42\ \text{hours}=1\ \text{hour}\ 25\ \text{minutes and }12\ \text{seconds}$

They will meet in $1\ \text{hour}\ 25\ \text{minutes and }12\ \text{seconds}$ .

wolverine [178]3 years ago

4 0

Answer: 10/7 hrs

Step-by-step explanation:

Step 1 of 5:Let’s organize the given information. Fill in the Rate column :

Rate (mph) Time (h) Distance (mi)

1st plane 40

2nd plane 100

Step 2 of 5:Let x hours be the time it will take them to meet. Complete the chart using the formula : Distance= Rate x Time :

Rate (mph) Time (h) Distance (mi)

1st plane 40 x

2nd plane 100 x

Step 3 of 5:We know that the distance between the planes was 200 miles.

Use this to make an equation : 40x+100x=200

.

Step 4 of 5:Solve the equation : 10/7

Step 5 of 5:Read the problem and answer the question in the problem.

It will take them

hours. 10/7 hrs

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10. In a survey of 212 people at the local track and field championship, 72% favored the home team

igomit [66]

Answer:

a. The margin of error for the survey is of 0.0308 = 3.08%.

b. The 95% confidence interval that is likely to contain the exact percent of all people who favor the home team winning is (65.96%, 78.04%).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

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The margin of error of the survey is:

$M = \sqrt{\frac{\pi(1-\pi)}{n}}$

The confidence interval can be written as:

$\pi \pm zM$

In a survey of 212 people at the local track and field championship, 72% favored the home team winning.

This means that $n = 212, \pi = 0.72$

a. Find the margin of error for the survey.

$M = \sqrt{\frac{0.72*0.28}{212}} = 0.0308$

The margin of error for the survey is of 0.0308 = 3.08%.

b. Give the 95% confidence interval that is likely to contain the exact percent of all people who favor the home team winning.

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Lower bound:

$\pi - zM = 0.72 - 1.96*0.0308 = 0.6596$

Upper bound:

$\pi + zM = 0.72 + 1.96*0.0308 = 0.7804$

As percent:

0.6596*100% = 65.96%

0.7804*100% = 78.04%.

The 95% confidence interval that is likely to contain the exact percent of all people who favor the home team winning is (65.96%, 78.04%).

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coldgirl [10]

Answer:

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Step-by-step explanation:

z-score:

In function of the margin of error M, the z-score is given by:

$Z = \frac{M}{\frac{\sigma}{\sqrt{n}}} = \frac{M\sqrt{n}}{\sigma}$

In this question, we have that:

$\sigma = 51.02, M = 2, Z = 1.96$

So

$Z = \frac{M\sqrt{n}}{\sigma}$

$1.96 = \frac{2\sqrt{n}}{51.02}$

$2\sqrt{n} = 51.02*1.96$

$\sqrt{n} = \frac{51.02*1.96}{2}$

$(\sqrt{n})^2 = (\frac{51.02*1.96}{2})^2$

$n = 2500$

The sample size required is 2500.

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