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koban [17]
3 years ago
11

Ms. Coleman's class has already collected 403 cans for a food drive, but the class wants to collect a total of at least 1,000 ca

ns. If each of the 28 students in the class brings an equal number of cans, how many more cans should each student bring for the class to have collected at least 1,000 cans?
Mathematics
2 answers:
makvit [3.9K]3 years ago
5 0

Answer:

Step-by-step explanation:

403 cans = current

1000 cans = goal

28 students

403 ÷ 28= 14.3- 14

1000÷ 28= 35.7- 36

∴ the students need to bring about 36 cans of food to be able to reach their goal

pochemuha3 years ago
4 0
We need to figure out how many more cans they need to bring.  We figure this out by subtracting 403 from 1,000, which gives us 596 more cans.  We now divide 596/28, to find how many each student would need to bring.  This equals 21.3.  You might want to mention that each student can't bring .3 of a can, so the answer could be 22 also.
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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

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

       E =  1.645* 10.24

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Generally the 90% confidence interval is mathematically represented as

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     -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

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    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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Step-by-step explanation:

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