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notka56 [123]
2 years ago
8

Three neon lights are turned on at the same time. One blinks every 4 seconds, the second one blinks every 5 seconds, and the thi

rd blinks every 6 seconds. In 5 minutes, how many times will all three lights blink at the same time?
Mathematics
1 answer:
emmasim [6.3K]2 years ago
4 0

In order to find how many times they will blink at the same time we need to find the common multiple of 4 and 6 between 1 and 60.

The first bulb blinks after every 4 seconds, means it blinks in multiple of 4.

The multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60.

The second bulb blinks after every 6 seconds, means it blinks in multiple of 6.

The multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.

The common multiple of 4 and 6 = 12, 24, 36, 48, 60.

∴

in sixty seconds they will blink 5 times at the same time.

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The p-value of the test is of 0.1922 > 0.02, which means that there is not significant evidence to reject the null hypothesis, that is, there is not significant evidence to conclude that the proportion is of less than 40%.

Step-by-step explanation:

Test if the proportion is less than 40%:

At the null hypothesis, we test if the proportion is of at least 0.4, that is:

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At the alternative hypothesis, we test if the proportion is of less than 0.4, that is:

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The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.4 is tested at the null hypothesis:

This means that \mu = 0.4, \sigma = \sqrt{0.4*0.6}

74 out of the 200 workers sampled said they would return to work

This means that n = 200, X = \frac{74}{200} = 0.37

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.37 - 0.4}{\frac{\sqrt{0.4*0.6}}{\sqrt{200}}}

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P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.37, which is the p-value of z = -0.87.

Looking at the z-table, z = -0.87 has a p-value of 0.1922.

The p-value of the test is of 0.1922 > 0.02, which means that there is not significant evidence to reject the null hypothesis, that is, there is not significant evidence to conclude that the proportion is of less than 40%.

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