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Hoochie [10]
3 years ago
6

2(n2 + 1) – 3(n2 – 3) + 3(2n2 – 2), n=(1/4)

Mathematics
1 answer:
IceJOKER [234]3 years ago
4 0

]Answer:

5\frac{5}{16}

Step-by-step explanation:

Substitute the value of the variable into the equation and simplify.

2(\frac{1}{4} ^{2} +1-3(\frac{1}{4} ^{2} -3)+3((2*\frac{1}{4}^{2}  )-2)

Solve it out.

Multiply / divide / add / subtract everything out.

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Which situation is represented by the equation below?
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Suppose that g(x) = f(x) + 6. Which statement best compares the graph of g(x) with the graph of f(x)?
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The shear strength of each of ten test spot welds is determined, yielding the following data (psi).
Mumz [18]

Answer:

a) Mean = 382.3 psi

Standard deviation = 20.8 psi

b) 392.7 psi

c) P(X<400)=0.8026

Step-by-step explanation:

a) The population mean can be estimated as equal to the mean of the sample, and the population standard deviation can be estimated from the sample standard deviation:

M=\frac{1}{n}\sum x_i=\frac{1}{10}(389+405+409+367+358+415+376+375+367+362)\\\\M=\frac{1}{10}(3823)=382.3\\\\\mu\approx M=382.3

s=\sqrt{\frac{1}{n-1}\sum (x_i-M)^2}=\sqrt{\frac{1}{10-1}(389-382.3)^2+...+(362-383.2)^2}\\\\\\s=\sqrt{\frac{1}{9}(3906.1)}=\sqrt{434}=20.8\\\\\\\sigma\approx s=20.8

b) We start by searching for the z-value for the 95th percentile. This value is

z=1.645:

P(z

Then, the strength value below which 95% of all welds will have their strengths is:

X=\mu+z\cdot \sigma/\sqrt{n}=382.3+1.645*20.8/\sqrt{10}\\\\X=382.3+32.9/3.2=382.3+10.4\\\\X=392.7

c) We calculate the probability of X being equal or less than 400 as:

z=(X-\mu)/\sigma=(400-382.3)/20.8=17.7/20.8=0.851\\\\\\P(X\leq400)=P(z

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How to write 2/5 as a decimal
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