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Aneli [31]
3 years ago
11

when the length of a rectangle is 5 and the width of it is x - 2 what are two expression to represent the perimeter

Mathematics
1 answer:
Jet001 [13]3 years ago
3 0

Answer:Perimeter=5*2+2*(x-2)

Step-by-step explanation:

Because perimeter =2*length+2*width

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What is the scale factor of the following dilation DO,K (9,6) → (3, 2)
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(9,6)

-------- = (3,2)

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This means that the dilation of (9, 6) to (3, 2) is (1/3(x), 1/3(y))

⭐ Answered by Hyperrspace (Ace) ⭐

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Solve for m 1/C+1/m=1/z
9966 [12]
Answer:  " m = zC / (C − z) " .
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Explanation:
_________________________
Given:  1/C + 1/m = 1/z ;  Solve for "m".

Subtract  "1/C" from each side of the equation:
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1/C + 1/m − 1/C = 1/z − 1/C  ;

to get:  1/m = 1/z − 1/C ;
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Now, multiply the ENTIRE EQUATION (both sides); by "(mzC"); to get ride of the fractions:
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mzC {1/m = 1/z − 1/C} ;

to get:  zC = mC − mz ;

Factor out an "m" on the "right-hand side" of the equation:

zC = m(C − z) ;  Divide EACH side of the equation by "(C − z)" ; to isolate "m" on one side of the equation;

zC / (C − z) = m(C − z) / m ;  to get:   24/8 = 3  24

zC/ (C − z) = m ;   ↔   m = zC/ (C − z) .
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3 years ago
A bookstore had 64 copies of a magazine. yesterday it sold 7/8 of them. how many copies were sold yesterday
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5 0
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Read 2 more answers
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Lostsunrise [7]

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Your answer is 120m4

Step-by-step explanation:

7 0
3 years ago
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A tobacco company claims that its best-selling cigarettes contain at most 40 mg of nicotine. The average nicotine content from a
tatuchka [14]

Answer:

t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722    

df=n-1=15-1=14  

p_v =P(t_{(14)}>2.722)=0.0083  

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

Step-by-step explanation:

Information provided

\bar X=42.6 represent the average nicotine content

s=3.7 represent the sample standard deviation

n=15 sample size  

\mu_o =40 represent the value to check

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic

p_v represent the p value for the test

System of hypothesis

We want to verify if the nicotine content of the cigarettes exceeds 40 mg , the system of hypothesis are:  

Null hypothesis:\mu \leq 40  

Alternative hypothesis:\mu > 40  

The statistic for this case would be:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

And replcing we got:

t=\frac{42.6-40}{\frac{3.7}{\sqrt{15}}}=2.722    

The degrees of freedom are:

df=n-1=15-1=14  

The p value would be:

p_v =P(t_{(14)}>2.722)=0.0083  

We see that the p value is lower than the significance level given of 0.01 so then we can conclude that the true mean for the nicotine content is significantly higher than 40 mg

6 0
3 years ago
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