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photoshop1234 [79]
3 years ago
11

Suppose a student reaches in the bag and randomly selects nine red chips and eleven yellow chips. Based on this sample, what is

a good estimate for the number of enrolled university students that are male?
Mathematics
1 answer:
nataly862011 [7]3 years ago
7 0

Answer:

There are more boys than girls. The red chips are girls are the boy are the yellow chips.  If more yellow chips were pulled out then there are more boys. 11 + 9 = 2. 10 + 9 = 19 + 1 = 20. The fraction of how many boys there are would be 11/21 and the fraction for girls would be 9/21.  Also 11 x 9 = 99. There would be 99 people if you times the number of girls and boys. But it's best to add because it would be in the smaller form if you were to times it.

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DaniilM [7]

Answer:

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Step-by-step explanation:

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Make g the subject of the formula <img src="https://tex.z-dn.net/?f=w%3D7-%5Csqrt%7Bg%7D" id="TexFormula1" title="w=7-\sqrt{g}"
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\sqrt{g} = 7-w

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Step-by-step explanation:

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3 years ago
Read 2 more answers
PLEASE HELP 7TH GRADE 100 PTS!!!
MariettaO [177]

f(x) = 5x is linear. Just a straight line with a slope of +5. So if the intervals are both a difference of 1, then the average rate of change will be the same.

f(x2) - f(x1) over x2 - x1. That's the formula for average rate of change.

So for Section A:

f(x) = 5x, (0,1)

[f(1) - f(0)]/(1-0)

= [5(1) - 5(0)]/1

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6 0
4 years ago
A national health organization warns that alcohol use among middle school students is on the rise. Concerned, a local health age
Lynna [10]

Answer:

CI 90 %  =  [ 0,05  ;  0,16 ]

Step-by-step explanation:

From sample we got

n  =  110

x  =  21

p =  21/110  =  0.19      then   q  =  1  -  p   q  =  1  -  0.19   q  = 0.81

To create a 90 % confidence Interval, the significance level α  = 10 %

α = 0.1    α/2  =  0.05

t(c) for   α/2  =  0.05    is from z-table    z(c) = 1.64

CI 90 %  =  (  p  ±   z(c) * √(p*q)/n

CI 90 %  =   [  0,11 ±  1.64 * √ (0.19*0.81)/110 ]

CI 90 %  =  [ 0.11  ±  1.64 * 0.037 ]

CI 90 %  =  [ 0.11  ±  0,06 ]

CI 90 %  =  [ 0,05  ;  0,16 ]

7 0
3 years ago
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