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777dan777 [17]
3 years ago
13

a quality control officer is randomly checki g the weights of flour bags being filled by an automstic filling machine. Each bag

is advertised as weighing 500 grams. A bag must weigh within 5.3grams in order to be accepted. What is the range of rejected bags x for the bags of flour​
Mathematics
1 answer:
gtnhenbr [62]3 years ago
7 0

Answer:

I think the correct answer from the choices listed above is the last option. The range of the rejected bags, x, for the bags of flour. A bag of flour is said to  weigh within 3.2 grams in order to be accepted. Therefore, it should weigh within 746.8 to 753.2 grams. Outside that range is rejected.

Step-by-step explanation:

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Can someone explain how to do #3?
dsp73

\displaysyle (\sqrt 2)^{3}=(2^{\frac{1}{2}})^3=2^{\frac{1}{2} \cdot 3}=2^{\frac{3}{2}}=\sqrt {2^3}=\sqrt 8

√8 is between 2 and 3, because 2²=4<8, but 3²=9>8. Also, our value is closer to 3 than to 2, so it is more than 2.5 and we have C and D options left.

Among these two numbers we find the one which is closer to √8.

C. 27=√729 ⇒ 2.7=√7.29

D. 28=√784 ⇒ <u>2.8=√7.84</u>

Hence our answer is D) 2.8

5 0
4 years ago
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Calculus 2 Master needed, show steps with partial fraction decomposition <img src="https://tex.z-dn.net/?f=%5Cint%283x%5E2-26x%2
Vladimir79 [104]

Answer:

-ln|x−5| + 2 ln(x²+4) + 3 tan⁻¹(x/2) + C

Step-by-step explanation:

The fraction will be split into a sum of two other fractions.

The first fraction will have a denominator of x − 5.  The numerator will the a polynomial of one less order, in this case, a constant A.

The second fraction will have a denominator of x² + 4.  The numerator will be Bx + C.

\frac{3x^{2}-26x+26}{(x-5)(x^{2}+4)}=\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}

Combine the two fractions back into one using the common denominator.

\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}=\frac{A(x^{2}+4)+(Bx+C)(x-5)}{(x-5)(x^{2}+4)}

This numerator will equal the original numerator.

A(x^{2}+4)+(Bx+C)(x-5)=3x^{2}-26x+26\\Ax^{2}+4A+Bx^{2}-5Bx+Cx-5C=3x^{2}-26x+26\\(A+B)x^{2}+(C-5B)x+(4A-5C)=3x^{2}-26x+26

Match the coefficients.

A+B=3\\C-5B=-26\\4A-5C=26

Solve the system of equations.

A=-1\\B=4\\C=-6

So we can rewrite the integral as:

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Solving:

\int {\frac{-1}{x-5}\, dx + \int {\frac{4x}{x^{2}+4}} \, dx - \int {\frac{6}{x^{2}+4} } \, dx

-\int {\frac{1}{x-5}\, dx + 2\int {\frac{2x}{x^{2}+4}} \, dx - 6\int {\frac{1}{x^{2}+4} } \, dx

-ln|x-5| + 2ln(x^{2}+4) - 6(\frac{1}{2} tan^{-1}(\frac{x}{2} )) + C

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3 years ago
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Step-by-step explanation:

The rectangle is: 7 * 5

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3 0
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