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german
3 years ago
9

The box plots show the student grades on a chapter test compared to the grades on a retest in the same class. Box plots titled C

hapter Test versus Retest with horizontal axis labeled Class Marks ranges from 0 to 100. Chapter Test box plot with minimum between 40 and 50 and maximum between 90 and 100 has interquartile ranges approximately between 50 and 80 and median approximately at 70. Retest box plot with minimum between 50 and 60 and maximum approximately at 100 has interquartile ranges approximately between 60 and 80 and median approximately between 60 and 70. Which of the following best describes the information about the medians? The chapter test and retest medians are almost the same. The chapter test and retest Q3 are the same, but the chapter test has the higher median. The chapter test median is much higher than the retest median. The low outlier on the chapter test pulls the median lower.
Mathematics
1 answer:
a_sh-v [17]3 years ago
8 0

Answer:

Step-by-step explanation:

C

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Help please ASAP!!!!!!!!!
Elden [556K]

Answer:

b. -33

Step-by-step explanation:

The question is about finding the determinant of a 2×2 matrix

General formula; when given matrix A =[  b  c ]

                                                                  [ d   e]

the determinant of A = (e×b)-(c×d)

Evaluate

(3  -3)

(-5  -6)  = (-6×3) - ( -5×-3)

            = (-18)- ( 15) = -33    

4 0
4 years ago
The average spending at Neco's salad bar is $9.03 with a standard deviation of $3.26. The distribution follows t-distribution. T
arlik [135]

Answer:

The difference between these cut-offs is of $1.9.

Step-by-step explanation:

In this question, we have to find the 90% confidence interval, using the t-distribution.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 41 - 1 = 40

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 40 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.95. So we have T = 1.684

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.684\frac{3.26}{\sqrt{41}} = 0.95

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 9.03 - 0.95 = $8.08.

The upper end of the interval is the sample mean added to M. So it is 9.03 + 0.95 = $9.98.

What will be the difference between the upper and lower spending cut-offs which define the middle 90% of the customers if the sample contains 41 customers

$9.98 - $8.08 = $1.9

The difference between these cut-offs is of $1.9.

8 0
3 years ago
Need help as fast as possible the question is 5(y+1)-y=4(y-1)+9​
erastovalidia [21]

Answer:

All real numbers are solutions.

4 0
3 years ago
Two angles in a triangle add up to 98 degrees . What is the size of the third angle ?
ira [324]
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3 years ago
You are going to mix 1 gallon bucket of window cleaner the instructions say to mix 1 part cleaner to 3 parts water how much clea
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3 years ago
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