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puteri [66]
2 years ago
11

LCM. Problem about how lcm works. picture in comments.

Mathematics
1 answer:
Ket [755]2 years ago
4 0
I can’t find a picture but i can explain how to find the LCM of two numbers.

Let’s say we want to find the LCM of 10 and 12.
We first write the prime factorization of both numbers:
10= 2x5

12= 2*2*3

We multiply the highest degree of the numbers:
That is, 2*2*3*5=60

https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-factors-and-multiples/cc-6th-lcm/v/least-common-multiple-exercise

You can also watch this video to get a better idea on how to find the LCM of two or more numbers.

Hope this helped.
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D = distance from finish line

D= 5 - 1/3x 

Distance from the finish line is equal to the distance of the start line from the finish line, minus one third of a km times the number of minutes hes been running.

using given variables; D= 5 - 1/3*3
D=5-3/3
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Plot (4, −34) on the coordinate plane.
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Step-by-step explanation:

Think about it. 4 is the x axis. -34 is the y axis. It will be located in quadrant IV.

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Leon throws a ball off a cliff. The graph represents a relation that models the ball’s height, h, in feet over time, t, in secon
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The Municipal Transit Authority wants to know if, on weekdays, more passengers ride the northbound blue line train towards the c
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Answer:

The 90% confidence interval  -49.8

The null hypothesis is  H_o : \mu_1 = \mu_2

The alternative hypothesis H_a  :  \mu_1 <  \mu_2

The distribution test statistics is t = -3.222

The rejection region is  p-value < \alpha

The decision rule is reject the null hypothesis

The conclusion is

      There is sufficient evidence to conclude that there are more passengers riding the 8:30  train

The p-value  is  p-value  =0.000951

Step-by-step explanation:

From the question we are told that

    The first sample size n_1 = 30

    The first sample mean is  \= x _1 = 323

    The first standard deviation is s_1 = 41

    The second sample size is n_2 = 45

    The second sample mean is  \=x_2 = 356

    The second standard deviation is s_2 = 45

given that the confidence level is 90% then the level of significance is mathematically represented as

          \alpha  = (100 -90)\%

         \alpha  = 0.10

Generally the critical value of \frac{\alpha }{2} obtained from the normal distribution table is  

   Z_{\frac{\alpha }{2} } = 1.645

Generally the pooled variance is mathematically represented as

        s^2 = \frac{(n_1 - 1)s_1^2  + (n_2 -1)s_2^2 }{n_1 + n_2 -2}

      s^2 = \frac{(30 -1)(41^2) + (45-1)45^2}{30+45 -2}

     s^2 = 1888.34

Generally the standard error is mathematically represented as

     SE =  \sqrt{\frac{s^2}{n_1} + \frac{s^2}{n_2}  }

=>  SE =  \sqrt{\frac{1888.34}{30} + \frac{1888.34}{45}  }

=>   SE =  10.24

Generally the margin of error is mathematically evaluated as  

      E =  Z_{\frac{\alpha }{2} } * SE

       E =  1.645* 10.24

       E = 16.85

Generally the 90% confidence interval is mathematically represented as

     \=x_1 -\=x_2 -E < \mu_1 -\mu_2 < \=x_1 -\=x_2 +E

     323 -356 -16.84

     -49.8

The null hypothesis is  H_o : \mu_1 = \mu_2

The alternative hypothesis H_a  :  \mu_1 <  \mu_2

Generally the test statistics is mathematically represented as

     t =  \frac{\= x_1 - \=x_2 }{SE}

=>   t = \frac{323-356}{10.24}

=>   t = -3.222

Generally the degree of freedom is mathematically represented as

     df =  n_1+n_2 -2

      df = 30 + 45 -2

     df = 73

The p-value is obtained from the student t distribution table at degree of freedom of 73 at 0.05 level of significance

    The value is  p-value  =0.000951

Here the level of significance is  \alpha =  5\%  =  0.05

Given that the p-value < \alpha then we  reject the null hypothesis

Then the conclusion is  

  There is sufficient evidence to conclude that there are more passengers riding the 8:30  train

               

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