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3 years ago

A train travels 95 kilometers in 4 hours, and then 94 kilometers in 3 hours. What is its average speed?

1 answer:

5 0

One way you could do this is by dividing 94 by 3 and then 95 by 4. This gives you the KPH (Kilometers per hour) of both trains. The first speed you mentioned, we will call that speed 'A.' The second one will be speed 'B.' Speed A's KPH is 23.75 MPH. Speed B's KPH is 31.333. We can change these into fractions (23 3/4 and 31 1/3.) These fractions could then be set into like terms. (This gives us 285/12 and 376/12) We then add these together. This gives us 661/12. The next thing to do is to convert this back to a decimal form, which gives us 55.0833.... Divide this number by two, and that is your answer. The average speed of the train is 29.04 KPH (This number is rounded to the nearest hundreth- be sure to make a note of that on your answer sheet)

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Give all relationships between angle 1 and angle 2 select all that apply

kolbaska11 [484]

Answer:

vertical maybe if try it and see

6 0

2 years ago

K(a)= 4a-4, Find k(-9)

FromTheMoon [43]

We simply replace a with -9

k(-9) = 4 * -9 - 4

k(-9) = -40

5 0

3 years ago

Find the median of the following data set.<br><br> 0.48, 0.66, 1.02, 0.82, 0.7, 0.94

kari74 [83]

Answer:

0.76

Step-by-step explanation:

0.48, 0.66, 1.02, 0.82, 0.7, 0.94

Organize the numbers from least to greatest.

0.48, 0.66, 0.7, 0.82, 0.94, 1.02

Then find the middle (Median is the middle).

0.48, 0.66, 0.7, |?| 0.82, 0.94, 1.02

Take the two numbers closest to the middle. Add them.

0.7 + 0.82

= 1.52

Divide that with 2.

= 0.76

8 0

3 years ago

83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi

arlik [135]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

Step-by-step explanation:

Information given

$\bar X=32.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom given by:

$df=n-1=8-1=7$

The confidence level is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.