The circumference of a circle is 87 cm. What is the DIAMETER of the circle?

HELP PLEASE!!!!!!! IM TIREDD AND AM STUCK ON THIS LAST QUESTION THAT IVE BEEN WORKIGN ON FOR 20 MINUTES

avanturin [10]

Answer:

1/3

Step-by-step explanation:

7 0

3 years ago

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.
<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

6 0

3 years ago

Answer please and thank you

Liono4ka [1.6K]

Answer:

b

a=1

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Carlos started a business with 8 customers. Each month, his business increased by 4 customers. He would like 52 customers a mont

maxonik [38]

Answer: 11 months

Step-by-step explanation:

Let the number of months that it will take him to reach his goal be represented by x.

Based on the information given in the question, this can be formed into an equation as:

8 + 4x = 52

4x = 52 - 8

4x = 44

x = 44/4

x = 11

It will take 11 months to reach the goal

5 0

3 years ago

Which of the following has a solution set of {x | x = 0}?

notka56 [123]

Answer:

(b) (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = {x | x = 0}

Step-by-step explanation:

Here, the given expression is : {x | x = 0}

So, the ONLY element in the given set = {0}

Now, take each option and solve the given expression:

(a) x + 1 < -1

Adding -1 BOTH sides, we get:

x + 1 -1 < -1 -1

or, x < - 2 ⇒ x = { -∞ , .... , -4,-3}

Also, x + 1 < 1

Adding -1 BOTH sides, we get:

x + 1 -1 < 1 -1

or, x <0 ⇒ x = { -∞ , .... , -4,-3,-2,-1}

So, (x + 1 < -1) ∩ (x + 1 < 1) = { -∞ , .... , -4,-3}∩ { -∞ , .... , -4,-3,-2,-1}

= { -∞ , .... , -4,-3}

⇒ (x + 1 < -1) ∩ (x + 1 < 1) ≠ {0}

Similarly, solving

(b) (x + 1 ≤ 1) ∩ (x + 1 ≥ 1)

(x + 1 ≤ 1) = x≤ 0 = { -∞ , .... , -4,-3,-2,-1, 0}

(x + 1 ≥ 1) = x ≥ 0 = {0,1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = {0}

(b)(x + 1 < 1) ∩ (x + 1 > 1)

(x + 1 < 1) = x < 0 = { -∞ , .... , -4,-3,-2,-1}

(x + 1 > 1) = x > 0 = { 1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = Ф

Hence, (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = {x | x = 0}

8 0

3 years ago

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