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MaRussiya [10]
3 years ago
14

Ben walks 500 meters from his house to the corner store. He then walks back toward his house, but continues 200 meters past his

house to talk to a neighbor. It takes Ben 17 minutes from the time he leaves his house until he stops to talk to his neighbor. What is Ben’s average velocity? Round to the nearest tenth.
Mathematics
1 answer:
lbvjy [14]3 years ago
8 0

Answer:1.2 m/s

Step-by-step explanation:

Total distance covered is 500+500+200 m so the average velocity is given by

V = distance/time

V=1200/(17*60)

V=1.176 m/sec

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Step-by-step explanation:

a) 2,3,4,5,6,7

b)4,6,7

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2 years ago
Which of the following is true about a nonlinear function?
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It is not straight and does not always pass through 0,0
so A, C, and D are incorrect.

Look up nonlinear function, and it shows a curved line.

The answer is B. It can be curved.
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The circumference of a circle is equal to n times its diameter. Estimate the circumference of a round swimming pool that has a d
sdas [7]

Answer:  An estimate done quickly: The circumference is about 75 ft.

Step-by-step explanation:

Round π to 3, and 24 to 25.  3 times 25 is 75

A more exact value would be 75.36

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5 0
3 years ago
A manager at the bookstore must arrange a collection of 231 books for display. The manager decides to use a combination of 15 sh
Kitty [74]

Answer:

6 tables.

Step-by-step explanation:

If he has 15 shelves, we can put the overall number of books off to the side for a moment. First calculate how many books he COULD put onto shelves. This would be 7*15, since each of fifteen shelves can hold 7 books. Counting by fifteens or using a calculator allows us to see that the shelves can hold 105 books.

Going back to the original number of books, we subtract 105 from the original value. The equation at this point is 231 - 105. The result is 126 books left to put on top of tables. But we're not done yet!

Since a table can hold 25 books, we need to divide the remaining number of books by 25. That would be 126/25. Doing this gives us 5 tables we would need, plus one book let over. Since we've run out of shelves, we MUST use another table just for the final book. That's 5+1, or 6 tables. Let's not forget to label our answer.

6 0
3 years ago
The quality control manager of a chemical company randomly sampled twenty 100-pound bags of fertilizer to estimate the variance
lisabon 2012 [21]

Answer:

The 95% confidence interval for the variance in the pounds of impurities would be 3.829 \leq \sigma^2 \leq 14.121.

Step-by-step explanation:

1) Data given and notation

s^2 =6.62 represent the sample variance

s=2.573 represent the sample standard deviation

\bar x represent the sample mean

n=20 the sample size

Confidence=95% or 0.95

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the 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 \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=20-1=19

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.025,19)" "=CHISQ.INV(0.975,19)". so for this case the critical values are:

\chi^2_{\alpha/2}=32.852

\chi^2_{1- \alpha/2}=8.907

And replacing into the formula for the interval we got:

\frac{(19)(6.62)}{32.852} \leq \sigma^2 \frac{(19)(6.62)}{8.907}

3.829 \leq \sigma^2 \leq 14.121

So the 95% confidence interval for the variance in the pounds of impurities would be 3.829 \leq \sigma^2 \leq 14.121.

4 0
3 years ago
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