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kenny6666 [7]
3 years ago
6

What percent of 45 is 36

Mathematics
1 answer:
Lina20 [59]3 years ago
5 0
For this question you can say: 
36/45 = ?/100
so ? is your answer:
100*36/45 = 80% :)))
i hope this is helpful
have a nice day 
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Triangle CDE is translated down and to the right forming triangle C prime de prime a prime which congruency statement is correct
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I have no idea, i just have to answer a question i’m sorry
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A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to fin
g100num [7]

Answer: (24.28,\ 27.72)

Step-by-step explanation:

Given : Sample size : n=50

Sample mean : \overline{x}=26

Standard deviation : \sigma =6.2

Significance level : \alpha=1-0.95=0.05

Critical value : z_{\alpha/2}=1.96

Formula to find the confidence interval for population mean :-

\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=26\pm(1.96)\dfrac{6.2}{\sqrt{50}}\\\\\approx26\pm1.72\\\\=(26-1.72,\ 26+1.72)\\\\=(24.28,\ 27.72)

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6 0
3 years ago
Tough one!
frosja888 [35]
Imx->0 (asin2x + b log(cosx))/x4 = 1/2 [0/0 form] ,applying L'Hospital rule ,we get

= > limx->0 (2a*sinx*cosx - (b /cosx)*sinx)/ 4x3 = 1/2 => limx->0 (a*sin2x - b*tanx)/ 4x3 = 1/2 [0/0 form],

applying L'Hospital rule again ,we get,

 = > limx->0 (2a*cos2x - b*sec2x) / 12x2 = 1/2

For above limit to exist,Numerator must be zero so that we get [0/0 form] & we can further proceed.

Hence 2a - b =0 => 2a = b ------(A)

limx->0 (b*cos2x - b*sec2x) / 12x2 = 1/2 [0/0 form], applying L'Hospital rule again ,we get,

= > limx->0 b*(-2sin2x - 2secx*secx.tanx) / 24x = 1/2 => limx->0 2b*[-sin2x - (1+tan2x)tanx] / 24x = 1/2

[0/0 form], applying L'Hospital rule again ,we get,

limx->0 2b*[-2cos2x - (sec2x+3tan2x*sec2x)] / 24 = 1/2 = > 2b[-2 -1] / 24 = 1/2 => -6b/24 = 1/2 => b = -2

from (A), we have , 2a = b => 2a = -2 => a = -1

Hence a =-1 & b = -2
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The plumber worked 4 hours total 
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Use the graph for this question what is the average rate of change from x=0 to x=2
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The rate of change should be 2
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