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Alina [70]
3 years ago
12

Which shows the mean, median, mode, and range for this set of data?

Mathematics
1 answer:
Brums [2.3K]3 years ago
6 0
8.3,4.5,6.2,9.6,4.3,11.2,5.9,9.7,10.5

Mean: 
To find mean you do the same as you would to find an average. Add up all of the numbers and divide by how many numbers are there. So, 
8.3 + 4.5 + 6.2 + 9.6 + 4.3 + 11.2 + 5.9 + 9.7 + 10.5 = 70.2 
Then divide by 9 since you have 9 numbers.
70.2 / 9 = 7.8 

Median:
To find median you set up the data from least to greatest and cross out one number from each side until you are left with one number as the middle number.
4.3, 4.5, 5.9, 6.2, 8.3, 9.6, 9.7, 10.5, 11.2 
       4.5, 5.9, 6.2, 8.3, 9.6, 9.7, 10.5
              5.9, 6.2, 8.3, 9.6, 9.7
                     6.2, 8.3, 9.6
                         = 8.3 
Mode:
Mode is the most, meaning the number that appears the most in your set of data. In this case there is no mode because in your set there is no number that occurs more than once.

Hope this helps!
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Someone could help me out to find the answer thanks​
Luden [163]

Answer:

  h(x) = 11/(x -4)

Step-by-step explanation:

The domain of m(x) is restricted to x≠1. That means the domain of m(n(x)) must be restricted so that n(x)≠1, or x≠4.

The only offered choice with a domain restriction x≠4 is the 3rd choice:

  h(x) = 11/(x -4)

4 0
3 years ago
Suppose an amount increases by 100%, then decreases by 100%. Find the final amount. Would the situation change if the original i
Anarel [89]
Well to solve this, lets plug in some numbers. Lets assume that for now, the starting number is 2. If it increases by 100%, bringing it up to 4, then decreases 100%, it is brought down to zero. Now one thing to note is that any number decrease by 100% is automatically zero. Watch, we start with two, it is then increased by 150%, bringing it up to 5, then decreased by 100%, bringing it back down to zero. In conclusion, 100% cannot be specified as a value, but rather a relation. 100% represents the entire number, meaning that if you are decreasing a number by 100%, you are basically subtracting the number by itself!
6 0
3 years ago
The Nellie Mae organization conducts an extensive annual study of credit card usage by college students. For their 2004 study, t
Masja [62]

Answer:

1. Null Hypothesis, H_0 : p_1-p_2=0  or  p_1=p_2  

  Alternate Hypothesis<u>,</u> H_A : p_1-p_2\neq 0  or  p_1\neq p_2

2. Test statistics = 4.63

    P-value = 0.00001

3. We conclude that the proportion of undergraduate students who held a credit card differed between these two years.

Step-by-step explanation:

We are given that the Nellie Mae organization conducts an extensive annual study of credit card usage by college students.

For their 2004 study, they analyzed credit bureau data for a random sample of 1,413 undergraduate students between the ages of 18 and 24. They found that 76% of the students sampled held a credit card. Three years earlier they had found that 83% of undergraduates sampled held a credit card.

<em>Let  </em>p_1<em> = population proportion of undergraduate students who held a credit card in year 2001</em>

<em />p_2<em> = population proportion of undergraduate students who held a credit card in year 2004</em>

1. <u>Null Hypothesis</u>, H_0 : p_1-p_2=0  or  p_1=p_2   {means that the proportion of undergraduate students who held a credit card does not differed between these two years}

<u>Alternate Hypothesis,</u> H_A : p_1-p_2\neq 0  or  p_1\neq p_2   {means that the proportion of undergraduate students who held a credit card differed between these two years}

The test statistics that will be used here is <u>Two-sample z proportion statistics</u>;

                        T.S.  = \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)

where, \hat p_1 = sample proportion of undergraduate students who held a credit card in 2001 = 83%

\hat p_2 = sample proportion of undergraduate students who held a credit card in 2004 = 76%

n_1 = sample of students surveyed in 2001 = 1,413

n_2 = sample of students surveyed in 2004 = 1,413

So, <em><u>test statistics</u></em>  =   \frac{(0.83-0.76)-(0)}{\sqrt{\frac{0.83(1-0.83)}{1,413} +\frac{0.76(1-0.76)}{1,413}} }  

                               =  4.63

2. <u><em>Hence, the value of test statistics is 4.63.</em></u>

Also, P-value is given by the following formula;

         P-value = P(Z > 4.63) = 1 - P(Z \leq 4.63)

                                             = 1 - 0.99999 = <u>0.00001</u>

<em />

3. <em>Since in the question we are not given the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical values between -1.96 and 1.96 for two-tailed test. Since our test statistics is does not lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.</em>

Therefore, we conclude that the proportion of undergraduate students who held a credit card differed between these two years.

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3 years ago
What percent of the total players scored 31-40 points during November
Tomtit [17]

Answer:

40%

Step-by-step explanation:

4÷10x100=40

As you have 10 players total and 4 players scored between 31 and 40 points

6 0
3 years ago
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Anestetic [448]

Answer:

a. $3000

Step-by-step explanation:


6 0
2 years ago
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