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Julli [10]
3 years ago
12

Guys pls help me im failing my grade and i really need help plsssss and dont do it for the points

Mathematics
1 answer:
Nataliya [291]3 years ago
6 0

Answer:

simplified:  \frac{3}{5}

unsimplified:  \frac{12}{20}

Step-by-step explanation:

Since the fractions have the same denominator, all we have to do is subtract the numbers in the numerator:

15 - 3 = 12

The difference of the two fractions is \frac{12}{20}. To simplify this answer, divide both the numerator and the denominator by 4:

12/4 = 3

20/4 = 5

The difference in simplified form is \frac{3}{5}.

I hope this helps. :)

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A driver's education course compared 1,500 students who had not taken the course with 1,850 students who had. Of those students
aivan3 [116]

Answer:

P-value is less than 0.999995 .

Step-by-step explanation:

We are given that a driver's education course compared 1,500 students who had not taken the course with 1,850 students who had.

Null Hypothesis, H_0 : p_1 = p_2 {means students who took the driver's education course and those who didn't took have same chances to pass the written driver's exam the first time}

Alternate Hypothesis, H_1 : p_1 > p_2 {means students who took the driver's education course were more likely to pass the written driver's exam the first time}

The test statistics used here will be two sample Binomial statistics i.e.;

                             \frac{(\hat p_1 - \hat p_2)-(p_1 - p_2)}{\sqrt{\frac{\hat p_1(1- \hat p_1)}{n_1} + \frac{\hat p_2(1- \hat p_2)}{n_2} } } ~ N(0,1)

Here, \hat p_1 = No. of students passed the exam ÷ No.of Students that had taken the course

 \hat p_1  = \frac{1150}{1850}            Similarly,  \hat p_2 = \frac{1440}{1500}          n_1 = 1,850       n_2 = 1,500

  Test Statistics = \frac{(\frac{1150}{1850} -\frac{1440}{1500})-0}{\sqrt{\frac{\frac{1150}{1850}(1- \frac{1150}{1850})}{1850} + \frac{\frac{1440}{1500}(1- \frac{1440}{1500})}{1500} } } = -27.38

P-value is given by, P(Z > -27.38) = 1 - P(Z > 27.38)

Now, in z table the highest critical value given is 4.4172 which corresponds to the probability value of 0.0005%. Since our test statistics is way higher than this so we can only say that the p-value for an appropriate hypothesis test is less than 0.999995 .

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3 years ago
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Since the bases are same now, powers can be added

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