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hjlf
3 years ago
8

Please Help Me!

Mathematics
2 answers:
Rainbow [258]3 years ago
4 0
For the first option, the range is a measure of variability which measures the spread of the data set from the least value to the greatest value, but it does not take into account the variability of the other data values of the data set. The range is easily affected by the presence of outliers (data points that are away from other data points). Thus the range is regarded as a weak measure of variability and is not used when other measures of variability are available. Thus, that the range of the two data sets are equal does not mean that the data sets have the same variability. Therefore, the first option is not the correct answer.

For the second option, the median is not a measure of variability. Thus, that a data set has a greater median than another data set does not mean that the data set would have a greater variability. Therefore, the second option is not the correct answer.
For the third option, the inter-quartile range (IQR) is a better measure of variability than the range because it takes into account more data points than the range. Now, because, the the IQR of Team 2 is less than the IQR of Team 1, this shows that Team 1 have greater variability than Team 2 and thus the conclusion of the coaches are inaccurate. Therefore, the third option is the correct answer.

For the fourth option, the mean absolute deviation, MAD, is a better measure of variability than the IQR because it takes into account all the points of the data set. While IQR measures variability with respect to the median, MAD measures variability with respect to the mean. Because we are told that the data sets are not symmetrical, the median will be a better measure of the center than the mean, thus the IQR will present a better measure of the variability of the data sets. Thus, though the MAD for Team 2 was calculated to be a larger number than the MAD for Team 1, the information can be misleading in arriving at a conclusion on which data set has more variability because the data sets are not symmetrical. Therefore, the fourth option is not the correct answer.
DochEvi [55]3 years ago
3 0

Answer:

It is the third option

Step-by-step explanation:

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5 0
3 years ago
The graph of the function f(x) = –(x + 6)(x + 2) is shown below.Which statement about the function is true?
LekaFEV [45]

Answer:

The first statement is true.

Step-by-step explanation:

The function is f(x) = - (x + 6)(x + 2)  

⇒ f(x) = - x² - 8x - 12

Now, condition for a function f(x) to be increasing at x = a is f'(a) > 0.

Now, f(x) = - x² - 8x - 12

⇒ f'(x) = -2x - 8 {Differentiating with respect to x}

Now, f'(a) = -2a - 8 {Here a can be any real value}

And, the condition for increasing function at x = a is  

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⇒ - 2a > 8

⇒ a < - 4

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4 0
3 years ago
A customer borrowed 2000 and then further 1000 both repayable in 12 months .What would he have saved if he had taken out one loa
creativ13 [48]

Answer:

No savings

Step-by-step explanation:

Let the interest rate be same for all the loans and consider it to be R.

<u>First case</u>

Principal 1 = 2000

Time = 12 moths = \frac{12}{12} = 1 year

Rate = R %

So, interest 1 = Principal * Time * Rate

                   = 2000 * 1 * R

                   = 2000R

Now,

Principal 2 = 1000

Time = 12 moths = \frac{12}{12} = 1 year

Rate = R %

So, interest 2 = Principal * Time * Rate

                   = 1000 * 1 * R

                   = 1000R

Total interest to be paid = 2000R + 1000R

                                         = 3000R

<u>Second case</u>

Principal = 3000

Time = 12 moths = \frac{12}{12} = 1 year

Rate = R %

Interest = Principal * Time * Rate

             = 3000 * 1 * R

             = 3000R

We conclude that there would be no savings because the interest to be paid in both the cases are same.

5 0
4 years ago
Read 2 more answers
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