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Murrr4er [49]
3 years ago
14

Can somebody help with this ?

Mathematics
1 answer:
Igoryamba3 years ago
8 0
You have to calculate the median and which the numbers all
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Hope an expert can help with this
worty [1.4K]

if there are 300 raffle tickets and one raffle ticket is drawn, the proabability of drawing the next one is 1 in 299 because there is one less ticket from the previous win

7 0
3 years ago
Please help I’ll mark you as brainliest if correct!
Pavlova-9 [17]

Answer:

3 ways

36÷6=6

36÷2=18

36÷3=12

maybe if not then 2 or 4

6 0
2 years ago
Read 2 more answers
Can someone please answer. There is one problem. There is a picture. Thank you!
jonny [76]
SA=4pir^2
78.54km^2=SA
78.54=4pir^2
divide both sides by 4
19.635=pir^2
aprox pi=3.14 so divide both sides by 3.14
6.25318=r^2
sqrt both sides
2.50063=r
round
radius=2.5km

this question is flawed in that it gives the wrong units

answer is the option with 2.5 in it
2nd one
5 0
3 years ago
I need help with math
Stels [109]
1/2 of an acre per hour
5 0
3 years ago
You use a line of best fit for a set of data to make a prediction about an unknown value. the correlation coeffecient is -0.833
alina1380 [7]

Answer: The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2  = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.

Step-by-step explanation: this is the same paragraph The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2  = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.

5 0
3 years ago
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