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saveliy_v [14]
3 years ago
10

Given the function f(x) = 5(x+4) − 6, solve for the inverse function when x = 19. (1 point) 1 68 72 84

Mathematics
1 answer:
vovangra [49]3 years ago
6 0

Answer:

x = 19, y = 1

Step-by-step explanation:

f(x) = 5(x + 4) - 6

y = 5(x + 4) - 6

inverse function:

switch x and y

x = 5(y + 4) - 6

solve for y

x = 5y + 20 - 6

x = 5y + 14

x - 5y = 14

-5y = -x + 14

y = (-x + 14)/-5

y = 1/5x - 2.8

plug 19 in for x

y = 1/5(19) - 2.8

y = 3.8 - 2.8

y = 1

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sladkih [1.3K]

Answer:

A. If the side lengths are the same, then a triangle is not scalene.

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6 0
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a class with 30 students had an average score of 80 on a test. A class with 20 students had an average score of 90 on the same t
snow_tiger [21]

The average score of all the students in both classes is 84.

<h3>How to calculate the average?</h3>

The class with 30 students had an average score of 80 on a test. The total score will be:

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A class with 20 students had an average score of 90 on the same test. The total score will be:

= 20 × 90

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8 0
1 year ago
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zubka84 [21]
Question 1a - You have got this correct, the median mark is 35
Question 1b- To work out the range, you must do the largest value subtract the smallest value. For your data, this would be 57 - 13 = 44

Question 2 - You have drawn the lines in the correct places, but you have not used a ruler. In order to get full marks on a question like this you must use a ruler.

Question 3 - Your lower quartile and median is correct, to find the upper quartile, we do 3(n/4). 'n' is the number of data points - in this case 120.
120/4 = 30
30 x 3 = 90
Go across the graph at 90 to see when the line is hit.
From what I can tell, the Upper Quartile is 3.

Next just find the maximum value.
Again, by the looks of it, the Maximum value is 8.5.

Now you have all the data needed for the box plot.

Min = 0
LQ = 0.8
Med = 2.1
UQ =  3
Max = 8.5

Using this information, draw a box plot like you did on the previous question.
<em>Again, I must stress that any line that you draw (unless purposely curved) must be drawn with a ruler; even on the graph at the top, during you working out. If you do not use a ruler, marks can be lost/taken away.</em>

Question 4a - For the median on a cumulative frequency chart, you must find the halfway point in the data. For this, we do n/2. In this case, n = 40
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Draw another one down to help you read the number.
The median looks like 34 seconds.
Question 4b - For this question, we already know the Min, Med and Max. Now we must work out the LQ and UQ.

Remember, LQ = n/4
In this case, n = 40
40/4 = 10
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To work out the UQ, we multiply the LQ place by 3 3(n/4).
3 x 10 = 30
30 on the graph takes us up to 45 seconds.

We now know that the:
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With this information, draw a box plot like you have in the previous questions.

Question 5 - Good things to always compare on box plots are the Min/Max/Range and the Inter Quartile Range.
The boys had a lowest minimum and a higher maximum, this means that their range is larger, resulting in a large spread of data in comparison to the girls.
The Inter Quartile Range is the difference between the Upper Quartile and the Lower Quartile (how wide the box is).
The boy IQR = 45 - 16 = 29
The girls IQR = 34 - 23 = 11
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Question 5a - The median in a box plot is always the line down the centre of the box. In this case:
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Question 5b - The IQR is always the UQ - LQ
With this data, the IQR is the following:
36 - 17 = 19 marks

Hope this helps
6 0
3 years ago
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