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Jet001 [13]
2 years ago
12

The sum of the first 100 positive whole numbers is 5050. What is the sum of the first 100 positive odd whole number?

Mathematics
1 answer:
Rufina [12.5K]2 years ago
4 0

The series of odd numbers is:

1, 3, 5, 7, 9, 11, 13, ....

If you observe this series, it is an Arithmetic Series with first term as 1 and common difference of 2. Common difference is defined as the difference between two consecutive terms of the series. So by using the formula of sum of an Arithmetic Series, we can find the sum of first 100 positive odd whole numbers.

The formula for the sum of an Arithmetic Series is:

S_{n}=\frac{n}{2}(2a_{1}+(n-1)*d)

Here,

n = number of terms = 100

a₁ = first term = 1

d = Common Difference = 2

Using the values, we get:

S_{100}=\frac{100}{2}(2(1)+99*2)\\   \\  S_{100}=10000

Thus, the sum of first 100 positive odd whole numbers is 10,000.

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400 students were randomly sampled from a large university, and 289 said they did not get enough sleep. Conduct a hypothesis tes
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Answer:

The sample proportion represents a statistically significant difference from 50%

Step-by-step explanation:

Null hypothesis: The sample proportion is the same as 50%

Alternate hypothesis: The sample proportion is not the same as 50%

z = (p' - p) ÷ sqrt[p(1 - p) ÷ n]

p' is sample proportion = 289/400 = 0.7225

p is population proportion = 50% = 0.5

n is number of students sampled = 400

z = (0.7225 - 0.5) ÷ sqrt[0.5(1 - 0.5) ÷ 400] = 0.2225 ÷ 0.025 = 8.9

The test is a two-tailed test. Using a 0.01 significance level, critical value is 2.576. The region of no rejection of the null hypothesis is -2.576 and 2.576.

Conclusion:

Reject the null hypothesis because the test statistic 8.9 falls outside the region bounded by the critical values -2.576 and 2.576.

There is sufficient evidence to conclude that the sample proportion represents a statistically significant difference from 50%.

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$24 saved after 3 weeks; $52 saved after 7 weeks. are these ratios equivalent
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To figure out this question, I first divided 24 into 3, which gives me an answer of 8. then I divided 52 into 7 , which also is 8 . because the two numbers are the same, I know the ratios are equivalent.
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Suppose you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by model
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Answer:

(1) The degrees of freedom for unequal variance test is (14, 11).

(2) The decision rule for the 0.01 significance level is;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
  • If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.  

(3) The value of the test statistic is 0.3796.

Step-by-step explanation:

We are given that you are an expert on the fashion industry and wish to gather information to compare the amount earned per month by models featuring Liz Claiborne's attire with those of Calvin Klein.

The following is the amount ($000) earned per month by a sample of 15 Claiborne models;

$3.5, $5.1, $5.2, $3.6, $5.0, $3.4, $5.3, $6.5, $4.8, $6.3, $5.8, $4.5, $6.3, $4.9, $4.2 .

The following is the amount ($000) earned by a sample of 12 Klein models;

$4.1, $2.5, $1.2, $3.5, $5.1, $2.3, $6.1, $1.2, $1.5, $1.3, $1.8, $2.1.

(1) As we know that for the unequal variance test, we use F-test. The degrees of freedom for the F-test is given by;

\text{F}_(_n__1-1, n_2-1_)

Here, n_1 = sample of 15 Claiborne models

         n_2 = sample of 12 Klein models

So, the degrees of freedom = (n_1-1, n_2-1) = (15 - 1, 12 - 1) = (14, 11)

(2) The decision rule for 0.01 significance level is given by;

  • If the value of our test statistics is less than the critical values of F at 0.01 level of significance, then we have insufficient evidence to reject our null hypothesis.      
  • If the value of our test statistics is more than the critical values of F at 0.01 level of significance, then we have sufficient evidence to reject our null hypothesis.  

(3) The test statistics that will be used here is F-test which is given by;

                          T.S. = \frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }  ~ \text{F}_(_n__1-1, n_2-1_)

where, s_1^{2} = sample variance of the Claiborne models data = \frac{\sum (X_i-\bar X)^{2} }{n_1-1} = 1.007

s_2^{2} = sample variance of the Klein models data = \frac{\sum (X_i-\bar X)^{2} }{n_2-1} = 2.653    

So, the test statistics =  \frac{1.007}{2.653 } \times 1  ~ \text{F}_(_1_4,_1_1_)

                                   = 0.3796

Hence, the value of the test statistic is 0.3796.

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Your correct answer is A. 0.6%

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