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Vikentia [17]
3 years ago
9

Mrs.Lewis bought 100 yards of ribbon.She used some of the ribbon to decorate dresses .Now she has 67 yards of ribbon remaining.

How many yards of ribbon did Mrs.Lewis use to decorate the dresses?
Mathematics
1 answer:
anastassius [24]3 years ago
5 0

Answer:

33 yards

Step-by-step explanation:

To find the number of ribbons that she used to decorate the dresses can be gotten by subtracting the amount of ribbon left from the total amount of ribbon she had.

She bought 100 yards. She has 67 yards left.

Therefore, the amount of ribbon she used for the dresses is:

100 - 67 = 33 yards

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3 years ago
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Integrate the following. ∫<img src="https://tex.z-dn.net/?f=5x%5E%7B4%7D%20dx" id="TexFormula1" title="5x^{4} dx" alt="5x^{4} dx
Vadim26 [7]

Answer:

\displaystyle D)  {x}^{5}  +  \rm C

Step-by-step explanation:

we would like to integrate the following Integral:

\displaystyle \int 5 {x}^{4} \, dx

well, to get the constant we can consider the following Integration rule:

\displaystyle \int c{x} ^{n}  \, dx  =  c\int  {x}^{n}  \, dx

therefore,

\displaystyle 5\int  {x}^{4} \, dx

recall exponent integration rule:

\displaystyle \int {x} ^{n}  \, dx  =  \frac{ {x}^{n + 1} }{n + 1}

so let,

  • n = 4

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\displaystyle  =  5\left( \frac{ {x}^{4+ 1} }{4 +  1}  \right)

simplify addition:

\displaystyle  =  5\left( \frac{ {x}^{5} }{5}  \right)

reduce fraction:

\displaystyle  =  {x}^{5}

finally we of course have to add the constant of integration:

\displaystyle  \boxed{ {x}^{5}  +  \rm C}

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The article "Occurrence and Distribution of Ammonium in Iowa Groundwater" (K. Schilling, Water Environment Research, 2002:177–18
Papessa [141]

Answer: 95% confidence interval for the difference between the proportions would be (1.31, 1.39).

Step-by-step explanation:

Since we have given that

Number of alluvial wells = 349

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Number of quaternary wells that had concentrations above 0.1 = 112

Average of alluvial wells = 0.27

Standard deviation = 0.4

Average of quaternary wells = 1.62

Standard deviation =1.70

So, 95% confidence interval gives

alpha = 5% level of significance.

\dfrac{\alpha}{2}=2.5\%\\\\z_{\frac{\alpha}{2}}=1.96

So, 95% confidence interval becomes,

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Hence, 95% confidence interval for the difference between the proportions would be (1.31, 1.39).

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