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3 years ago

15

A passenger train left Chicago and traveled toward Elkhart at an average speed of 29 mph. A freight train left 2.4 hours later a

nd traveled in the same direction but with an average speed of 36.5 mph. How long did the passenger train travel before the freight train caught up? *

1 answer:

ElenaW [278]3 years ago

5 0

29*2.4-36.5= 33.1 hrs before the freight train cought up.

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For the following exercises, solve each inequality and write the solution in interval notation.

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Answer:

The solution of the given set in interval form is $(-\infty,-4] \cup[12, \infty)$ .

Step-by-step explanation:

It is given in the question an inequality as $|x-4| \geq 8$ .

It is required to determine the solution of the inequality.

To determine the solution of the inequality, solve the inequality $x-4 \geq 8$ and, $x-4 \leq-8$

Step 1 of 2

Solve the inequality $x-4 \geq 8$

$\begin{aligned}&x-4 \geq 8 \\&x-4+4 \geq 8+4 \\&x \geq 12\end{aligned}$

Solve the inequality $x-4 \leq-8$ .

$\begin{aligned}&x-4 \leq-8 \\&x-4+4 \leq-8+4 \\&x \leq-4\end{aligned}$

Step 2 of 2

The common solution from the above two solutions is x less than -4 and $x \geq 12$ .

The solution set in terms of interval is $(-\infty,-4] \cup[12, \infty)$ .

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1 year ago

Construct the confidence interval for the population mean mu. c = 0.90, x = 16.9, s = 9.0, and n = 45. A 90% confidence inte

Georgia [21]

Answer:

The 90% confidence interval for population mean is $14.7 < \mu < 19.1$

Step-by-step explanation:

From the question we are told that

The sample mean is $\= x = 16.9$

The confidence level is $C = 0.90$

The sample size is $n = 45$

The standard deviation

Now given that the confidence level is 0.90 the level of significance is mathematically evaluated as

$\alpha = 1-0.90$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the standardized normal distribution table. The values is $Z_{\frac{\alpha }{2} } = 1.645$

The reason we are obtaining critical values for $\frac{\alpha }{2}$ instead of that of $\alpha$ is because $\alpha$ represents the area under the normal curve where the confidence level 1 - $\alpha$ (90%) did not cover which include both the left and right tail while $\frac{\alpha }{2}$ is just considering the area of one tail which is what we required calculate the margin of error

Generally the margin of error is mathematically evaluated as

$MOE = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

substituting values

$MOE = 1.645* \frac{ 9 }{\sqrt{45} }$

$MOE = 2.207$

The 90% confidence level interval is mathematically represented as

$\= x - MOE < \mu < \= x + MOE$

substituting values

$16.9 - 2.207 < \mu < 16.9 + 2.207$

$16.9 - 2.207 < \mu < 16.9 + 2.207$

$14.7 < \mu < 19.1$

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During statistical analysis, the investigator may consider searching for statistical associations among various groups that may

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Answer:

This is known as Data dredging.

Step-by-step explanation:

In order for any statistical analysis to be significant, a researcher has to first generate a hypothesis.

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In this case any correlation (relationship between cause-effect)will be purely by chance.

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