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emmainna [20.7K]
3 years ago
12

HELP ME PLZ HELP ME PLZ HELP ME PLZ I REALLY NEED YOUR help

Mathematics
2 answers:
arsen [322]3 years ago
8 0
Question 9 answer is 80 and for 10 it is 72 and for 11 it is 112 and for 12 it is 143

USPshnik [31]3 years ago
5 0

Answer:

Question 9= 80

Question 10= 72

Question 11= 112

Question 12= 143

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Can someone please explain to me how to find the interquartile range of something?
Alika [10]
I’ll do an example you have the numbers 1,5,7,2,9 you can tell the median is 7 right? So go from the very left and the median (7) and do the same thing you do to find median. (Id recommend crossing the numbers off as you go if that makes sense! :)
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3 years ago
What is the relationship between the 6s in 66,0472
vagabundo [1.1K]
They are both in the thousands places
5 0
3 years ago
What is the difference between 12/8 and 3/4?
Katen [24]

Answer:

3/4

Step-by-step explanation:

Difference means subtraction

12/8 - 3/4

We need a common denominator of 8

12/8 - 3/4 * 2/2

12/8  - 6/8

 6/8

Divide by 2 in the numerator and denominator

 3/4

6 0
3 years ago
Read 2 more answers
Solve this the link is down below
kogti [31]
There is no link here
4 0
3 years ago
There are currently 17 frogs in a (large) pond. The frog population grows exponentially, tripling every 6 days. How long will it
Tema [17]

Answer:

t=11,11 days

Step-by-step explanation:

F=frogs poblation, t=time, be the variables dF/dt = KF, dF/F=Kdt, integrating \int\limits^ {} \, dF/F =K\int\limits^ {} \, dt⇒ LnF=Kt+c,F=ce^{Kt}; Knowing t=0, F=17 and t=6 F=51 (tripling every 6 days (17*3)), F=ce^{K0} = F=c=17; F=17e^{6K} =51⇒e^{6K} =51/17; K6=ln\frac{51}{17} ; K=ln\frac{51}{17}/6=0.183, so F=17e^{0,183t}, now if F=130, t=? we have:

130=17e^{0.183t} =e^{0.183t} =130/17; 0.183t=ln(130/17); t=ln(130/17)/0.183 = 11,11

8 0
4 years ago
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