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3 years ago

{20,1,6,10,11} Find the subset and the proper subset.

1 answer:

4 0

With 5 elements in A={20,1,6,10,11}, there are 2^5=32 possible subsets, including
the null set, and A itself.
Any subset that is identical to A is NOT a proper subset.
Therefore there are 31 proper subsets, plus the subset {20,1,6,10,11}.

The subsets are:
null set {} (has no elements) ........total 1
{20},{1},{6},{10},{11}.......................total 5
{20,1},{20,6}...{10,11}.....................total 10
{20,1,6},{20,1,10},...{6,10,11}.........total 10
{20,1,6,10}...{1,6,10,11}.................total 5
{20,1,6,10,11}.................................total 1
Altogether 32 subsets.

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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 .

6 0

3 years ago

2n+5=20?????????????

Oksanka [162]

2n+5=20

Subtract 5 from 5 and 20

2n=15

Divide 15 and 2 by 2

n=7.5

8 0

3 years ago

Read 2 more answers

There are five consecutive even integers. The sum of the first two is thirty-six less than the sum of the last three numbers. Wh

Norma-Jean [14]

5 consecutive even integers: 2n, 2n+2, 2n+4, 2n+6, 2n+8 for any integer n The sum of the two smallest of five consecutive even integers is 50 less than the sum of the other three integers: 2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 504n + 2 = 6n -3234 = 2nn = 17 <span>The smallest integer in our sequence is 2n = 2*17 = 34</span> Check:2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 5034 + 36 = 38 + 40 + 42 - 5070 = 120 - 5070 = 70

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3 years ago

A molasses can is made with a cylindrical base having a radius of 6 cm, that is topped by a smaller cylindrical pouring spout wi

wariber [46]

Answer:

The total height of the can is 20cm

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2 years ago

A recent survey was conducted to compare the cost of solar energy to the cost of gas or electric energy. Results of the survey r

Mariana [72]

Answer:

c) at most 11.1%

Step-by-step explanation:

We have the data that is 97 ± 12, with 97 being the mean and 12 the standard deviation.

Now, the percentage of people who reached them for less than 73 dollars, if it were a normal distribution:

z = (73 - 97) / 12 = - 2

so it would be, a probability of 0.0228 or 0.228%.

But we don't know what distribution it has, but we can get an idea.

A and D discarded, as they are very high values, and 73 is well below the average.

B) is still a very high value.

Therefore the answer is C, at most 11.1%

4 0

3 years ago