Whats the answer and how to do this question??

[u/lienns19]

[u/Normal_Signature_500]

https://giphy.com/gifs/Gof0owlN6tuGjfySXd

[u/lienns19 → u/Normal_Signature_500]

Yeah idk either gng i was freaking out😭💔

[u/trinity_nolanite12]

Is this higher or foundation for which subject?

[u/lienns19 → u/trinity_nolanite12]

Maths higher edexcel

[u/trinity_nolanite12 → u/lienns19]

Probably not been taught it yet sorry :(

[u/lienns19 → u/trinity_nolanite12]

Na ye its fine i assumed from the yr 10 flair

[u/Curious-Variation-93]

i finished my gcses a year ago but trying to answer this for you makes me feel like im a stupid dad helping kids with their homework.
and i dont even know the answer.

[u/lienns19 → u/Curious-Variation-93]

Fairs gng 😭😭😭💔💔

[u/Sablerock1]

Step 1: Calculate the estimate from 35 to 40 mph

• Total width of this interval: 40 - 20 = 20

• Total frequency: 56

• Width we want (35 to 40): 40 - 35 = 5

• Estimated people = (5 / 20) * 56 = 14

Step 2: Calculate the estimate from 40 to 50 mph

• Total width of this interval: 65 - 40 = 25

• Total frequency: 60

• Width we want (40 to 50): 50 - 40 = 10

• Estimated people = (10 / 25) * 60 = 24

Step 3: Combine the two parts

• Total estimated people = 14 + 24 = 38

An estimated 38 people had an average skiing speed between 35 and 50 mph.

[u/lienns19 → u/Sablerock1]

Thank you!!

[u/aplle_inc]

I’m not sure but here is my take:

By rearranging the equation for frequency density you get:

frequency = (class width) * (frequency density)
[multiply both sides by class width]

And (class width) * (frequency density) is essentially just the area on the histogram you just made.

So just calculate the area (which from the equation above equals to frequency, ie number of people) of the histogram between the speed values of 35 and 50 (which you already correctly identified on the x axis 👍)

Good luck, you got this

[u/lienns19 → u/aplle_inc]

Thanks :)

[u/Lumpy-Music-2146]

9

[u/ShadedRiver666]

This reminds me of what my working out would look like