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kondaur [170]
2 years ago
6

Emergency, pls answerrrrr

Mathematics
2 answers:
ASHA 777 [7]2 years ago
8 0
3/10 because im a different breed
IrinaVladis [17]2 years ago
7 0

Answer:

Let the total fraction =1

Fraction eaten =½( morning),x(night)

Now,1-(½+x)=1/5

multiply through by LCM 5

5-5/2-5x=1

multiply through again by 2

10-5-10x=2

5-10x=2

5-2=10x

3=10x

x=3/10

So Fluffy ate 3/10 at night

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If you know the answers pls put this <br> 1<br> 2<br> 3<br> 4<br> 5
Hitman42 [59]

Answer:

1.6 kids

2.6

3.6

4.6

5.6

Step-by-step explanation:

if you count by 6 then you will see that im right

4 0
2 years ago
Find The distance between (-3,1) and (-1,6). If necessary,round to the nearest tenth.
kifflom [539]
Use distance formula:-

distance =  sqrt  [ (x2 - x1)^2 + (y2 - y1)^2) ]

let (x1,y1) = (-3,1) and (x2,y2) = (-1,6)   Plug these values in to formula.
8 0
3 years ago
Which of these limits evaluate to 0?
Vikentia [17]
<h3>Answer: C) I and II only</h3>

===============================================

Work Shown:

Part I

\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{2-2}{2+2}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = \frac{0}{4}\\\\\\\displaystyle \lim_{x \to 2}\frac{x-2}{x+2} = 0\\\\\\

----------

Part II

\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{\sin(0)}{0+2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = \frac{0}{2}\\\\\\\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x+2} = 0\\\\\\

----------

Part III

\displaystyle \lim_{x \to 5}\frac{x}{x} = \lim_{x \to 5}1\\\\\\\displaystyle \lim_{x \to 5}\frac{x}{x} = 1\\\\\\

7 0
2 years ago
What is the measurement of 3/4 in simplest form
stepan [7]

Answer:

It is in simplest form

Step-by-step explanation:

6 0
3 years ago
Construct a 99% confidence interval for the population standard deviation of white blood cell count (in cells per microliter). A
kogti [31]

Answer:

The correct option is 1.148 < σ < 6.015

Explanation:

The 99% confidence interval for the standard deviation is given below:

\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\frac{0.01}{2} }} }

Where:

\chi^{2}_{\frac{0.01}{2} }=18.548

\chi^{2}_{1-\frac{0.01}{2}}=0.676

Therefore, the 99% confidence interval is:

\sqrt{\frac{(7-1)2.019^{2}}{18.548} }

1.148

Therefore, the option 1.148 < σ < 6.015 is correct

3 0
3 years ago
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