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3 years ago

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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?

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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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

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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

$df = n_1+n_2 -2$

$df = 30 + 45 -2$

$df = 73$

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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