What is the reduced fraction of 25 and 60?

Could someone help with the questions in the images below? I found them difficult

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.
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.

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

6 0

4 years ago

A strand of bacteria has a doubling time of 15 minutes. If the population starts with 10 organisms, how long would it take for t

slamgirl [31]

In 90 minutes the population to grow to 700 organisms.

Given that,

A strand of bacteria has a doubling time of 15 minutes.

If the population starts with 10 organisms.

We have to determine,

How long would it take for the population to grow to 700 organisms?

According to the question,

A strand of bacteria has a doubling time of 15 minutes.

If the population starts with 10 organisms.

In 15 minutes strand of bacteria has a doubling time the population starts with 10 organisms.

$\rm = 10 \times 2 = 20 \ bacteria$

In 15 minutes 20 bacteria for the population to grow.

In 30 minutes strand of bacteria has a doubling time the population starts with 20 organisms.

$\rm = 20 \times 2 = 40 \ bacteria$

In 30 minutes 40 bacteria for the population to grow.

In 45 minutes strand of bacteria has a doubling time the population starts with 40 organisms.

$\rm = 40 \times 2 = 80 \ bacteria$

In 45 minutes 80 bacteria for the population to grow.

In 60 minutes strand of bacteria has a doubling time the population starts with 80 organisms.

$\rm = 80 \times 2 = 160 \ bacteria$

In 60 minutes 160 bacteria for the population to grow.

In 75 minutes strand of bacteria has a doubling time the population starts with 160 organisms.

$\rm = 160 \times 2 = 320 \ bacteria$

In 75 minutes 320 bacteria for the population to grow.

In 90 minutes strand of bacteria has a doubling time the population starts with 320 organisms.

$\rm = 320 \times 2 = 640 \ bacteria$

In 90 minutes 640 bacteria for the population to grow.

In 120 minutes strand of bacteria has a doubling time the population starts with 640 organisms.

$\rm = 640 \times 2 = 1280 \ bacteria$

In 120 minutes 1280 bacteria for the population to grow.

Hence, In 90 minutes the population to grow to 700 organisms.

For more details refer to the link given below.

brainly.com/question/3188472

3 0

2 years ago

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The frequency table categorizes winners of a prestigious award from.1990-2012 by their age at the time they received the award.

Ivan

Answer:

is there an image of the frequency table attached? whats the question

Step-by-step explanation:

3 0

3 years ago

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Please help! ill give brainliest!

prohojiy [21]

B

Step by step equation

7 0

3 years ago

How many solutions are there to the equation below?

Vikentia [17]

Answer:

$5x + 48 + 7x = 12(x + 4) \\ 5x + 48 + 7x = 12x + 48 \\ (5x + 7x) + 48 = 12x + 48 \\ 12x + 48 = 12x + 48 \\ 12x - 12x = 48 - 48 \\ 0 = 0$

There is no solution

4 0

3 years ago

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