Answer:
y = 2x+3
Step-by-step explanation:
Sorry I'm probably too late since Quizziz doesn't wait unless you have the freeze timer thing.
y = mx+b
b is the y intercept. The y intercept (when x=0) is given in the table.
Now plug a coordinate in to solve for m.
My coordinates: (1,5)
5 = m + 3
Solve.
2 = m
m = 2
Note: never plug in the y intercept when looking for m.
You can also find the slope by (subtracting the y value of one coordinate from another y value of another coordinate)/(subtracting x value from the first y coordinate above from the x value from the second coordinate).
In this case, it'd be
m = (7-5)/(2-1)
m = 2/1
m = 2
Problem 1
<h3>Answer: False</h3>
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Explanation:
The notation (f o g)(x) means f( g(x) ). Here g(x) is the inner function.
So,
f(x) = x+1
f( g(x) ) = g(x) + 1 .... replace every x with g(x)
f( g(x) ) = 6x+1 ... plug in g(x) = 6x
(f o g)(x) = 6x+1
Now let's flip things around
g(x) = 6x
g( f(x) ) = 6*( f(x) ) .... replace every x with f(x)
g( f(x) ) = 6(x+1) .... plug in f(x) = x+1
g( f(x) ) = 6x+6
(g o f)(x) = 6x+6
This shows that (f o g)(x) = (g o f)(x) is a false equation for the given f(x) and g(x) functions.
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Problem 2
<h3>Answer: True</h3>
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Explanation:
Let's say that g(x) produced a number that wasn't in the domain of f(x). This would mean that f( g(x) ) would be undefined.
For example, let
f(x) = 1/(x+2)
g(x) = -2
The g(x) function will always produce the output -2 regardless of what the input x is. Feeding that -2 output into f(x) leads to 1/(x+2) = 1/(-2+2) = 1/0 which is undefined.
So it's important that the outputs of g(x) line up with the domain of f(x). Outputs of g(x) must be valid inputs of f(x).
The muffin is 1.25
The bagel is 3.50
You get this by seeing the first person bought 3 muffins and 1 bagel.
The second vought only 1 muffin less, so subtract the 2 totals and that gives you the difference which is the price of the muffin
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
40/2 = 20
Draw a line (using ruler) across at 20 until you reach the line.
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
Look across at ten and you will find that the LQ = 16 seconds
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:
Min = 9
LQ = 16
Med = 34
UQ = 45
Max = 57
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
Again, the girls have much more concise results; even though they did not get the quickest result, they were more like one another.
Question 5a - The median in a box plot is always the line down the centre of the box. In this case:
Median = 24 marks
Question 5b - The IQR is always the UQ - LQ
With this data, the IQR is the following:
36 - 17 = 19 marks
Hope this helps
Answer:
we can not reject any value
Step-by-step explanation: From data we can test the highest and the lowest value to evaluate if one of these values are out of certain confidence Interval
If we established CI = 95 % then α = 5 % and α/2 = 0,025
From data we find the mean of the values
μ₀ = 12,03 and σ = 0,07
From z table we find z score for 0,025 is z(c) = ± 1,96
So limits of our CI are:
12,03 + 1,96 = 13,99
12,03 - 1,96 = 10,07
And all our values are within ( 10,07 , 13,99)
So we can not reject any value
