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Ratling [72]
3 years ago
14

Three pigs entered a race around a track. Piggly takes 12 minutes to run one lap. Piglet takes 9 minutes to run one lap. And it

takes wiggly 16 minutes to run one lap. If all three pigs bevin the race at the same time, how many minutes will it take for all three pigs cross the finish line at the same time
Mathematics
1 answer:
Korolek [52]3 years ago
7 0

144 Minutes


The least common multiple for 9 12 and 16 is 144

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The frequency table categorizes winners of a prestigious award from.1990-2012 by their age at the time they received the award.
Ivan

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is there an image of the frequency table attached? whats the question

Step-by-step explanation:

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The slope between the points (-1,-5) and (r, -9) is 2. What is the value of r?
Gnom [1K]

Answer:

r = 2

Step-by-step explanation:

slope = (y2 - y1)/(x2 - x1)

(-1, -5)

(r, -9)

-5 - (-9) / -1 - r = 2

4 / -1 - r = 2

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Please let me know if I've gotten anything wrong!

4 0
2 years ago
A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mat
Helen [10]

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be 8.400 \leq \sigma^2 \leq 39.827.

Step-by-step explanation:

1) Data given and notation

s^2 =16 represent the sample variance

s=4 represent the sample standard deviation

\bar x represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

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 \leq \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 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.01,19)" "=CHISQ.INV(0.99,19)". so for this case the critical values are:

\chi^2_{\alpha/2}=36.191

\chi^2_{1- \alpha/2}=7.633

And replacing into the formula for the interval we got:

\frac{(19)(16)}{36.191} \leq \sigma^2 \leq \frac{(19)(16)}{7.633}

8.400 \leq \sigma^2 \leq 39.827

So the 98% confidence interval for the variance in the pounds of impurities would be 8.400 \leq \sigma^2 \leq 39.827.

7 0
3 years ago
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