The ratio of girls to boys in Jaime's chorus is 2 to 4. If there are 25 girls in her chorus, how many boys are there? *

Nostrana [21]

height or altitude for a parallelogram means the perpendicular distance from one parallel side to the other. Check the picture below.

7 0

2 years ago

That's the question ??

Oksanka [162]

Step-by-step explanation:

Before we answer, the area of a Circle given it radius is

$a = \pi r {}^{2}$

The area of a circle given it diameter is

$a = \pi( \frac{d}{2} ) {}^{2}$

So let answers question 1-4.

1. The area is

$1024\pi$

2. The area is

$576\pi$

3.

$1.225\pi$

4.

$309.76\pi$

For the 5th problem, circumference is

$\pi \times d$

where d is the diameter and

$2\pi r$

where r is the radius.

A circumference of 100 m means the radius is 50 because

$2\pi \times r = 100\pi$

$r = 50$

So this means the area is

$2500 {\pi}$

3 0

2 years ago

A teacher randomly chooses a two-person leadership team from a group of four qualified students. Three of the students, Sandra,

nata0808 [166]

Answer:

What is P(A), the probability that the first student is a girl? (3/4)

What is P(A), the probability that the first student is a girl? (3/4)What is P(B), the probability that the second student is a girl? (3/4)

What is P(A), the probability that the first student is a girl? (3/4)What is P(B), the probability that the second student is a girl? (3/4)What is P(A and B), the probability that the first student is a girl and the second student is a girl? (1/2)

The probability that the first student is a girl is (3/4), likewise for the 2nd 3rd and 4th it's still (3/4). The order you pick them doesn't matter.

However, once you're looking at P(A and B) then you're fixing the first position and saying if the first student is a girl what's the probability of the second student being a girl.

3 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

Plz plz plz help me with this question

Gnoma [55]

My teacher told be to try and think of it as a circle, because circles are 360 degrees. Then to place those angles in the circle and it should make it easier to find the angles, because when you see it in the circle, you can divide it into 4 parts so that you would then have 4, 90 degree angles that you can then work off of by placing those angles you need to find into the circle. It was supposed to make it easier, and i genuinely hope that you can do this. You got this, and you will get through the pain of math tonight and many more. Good luck. ^-^

8 0

2 years ago

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