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Westkost [7]
3 years ago
15

Phil strung two sets of lights around the window of his store. The first set of lights blinks every 30s and the second every 45s

. How often do the sets of lights blink at the same time?
Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
5 0

Answer:

Every 90 seconds, the two sets of light blink together.

Step-by-step explanation:

Given that there are two sets of lights.

First light blinks at every 30s and

Second light blinks at every 45s

To find:

How often the sets of lights blink together?

Solution:

As per given statement,

First light blinks after 30s, 60s, <u>90s</u>, 120s, 150s, <u>180s</u>, ........

Second light blinks after 45s, <u>90s</u>, 135s, <u>180s</u>, .....

If we look at the common time, shown by underlining it, the two sets of light blink at the same time after 90s, 180s, ...... and so on.

So, the two lights blink together after every 90s.

This can also be solved by finding LCM (Least Common Multiple) of the two numbers i.e. 30 and 45.

Let us have a look at the factors of 30 and 45.

30 = 2 \times <u>3</u> \times <u>5</u>

45 = 3 \times <u>3</u> \times <u>5</u>

<u />

So, LCM = 2 \times 3 \times 3 \times 5 = <em>90</em>

<em></em>

So, after every <em>90s, </em>the lights will blink together.

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And we can use the complement rule and we got:

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Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the length of time waiting to be seated of a population, and for this case we know the distribution for X is given by:

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Part a

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P(X >20| X>15)

And if we use the conditional probability formula we got:

P(X >20| X>15)= \frac{P(X >20 \cap X>15)}{P(X>15)}=\frac{P(X>20)}{P(X>15)}

We can solve the problem using the z score formula given by:

z = \frac{x-\mu}{\sigma}

z_1 =\frac{20-19}{4} =0.25

z_2 =\frac{15-19}{4} =-1

And we can use the complement rule and we got:

P(z>0.25)=1-P(z

P(z>-1)=1-P(z

And replacing we got:

P(X >20| X>15)= \frac{0.401}{0.841}= 0.477

Part b

P(X >20| X>18)= \frac{P(X >20 \cap X>18)}{P(X>18)}=\frac{P(X>20)}{P(X>18)}

We can solve the problem using the z score formula given by:

z = \frac{x-\mu}{\sigma}

z_1 =\frac{20-19}{4} =0.25

z_2 =\frac{18-19}{4} =-0.25

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And replacing we got:

P(X >20| X>18)= \frac{0.401}{0.599}= 0.669

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