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4 years ago

The area of a gymnasium floor is 264 square yards the floor is 11 yards wide how long is the floor

Mathematics

2 answers:

dolphi86 [110]4 years ago

5 0

264/ 7 yes trust meeeeeeeeeeeeeeeeeeeeeeeee

Arisa [49]4 years ago

3 0

The equation to find area is:
$A=lw$
A is your area, l is your length, and w is your width. Let's insert the numbers we already have, which are our area and our width.
$264=11l$
To solve this, we must divide both 11l and 264 by 11, to get our variable l all alone on one side.
264 / 11 = ?
Solve this, and you'll have your length :)

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When an amount of heat Q (in kcal) is added to a unit mass (kg) of a

lisov135 [29]

Answer:

The change in temperature per minute for the sample, dT/dt is 71. $\overline {6}$ °C/min

Step-by-step explanation:

The given parameters of the question are;

The specific heat capacity for glass, dQ/dT = 0.18 (kcal/°C)

The heat transfer rate for 1 kg of glass at 20.0 °C, dQ/dt = 12.9 kcal/min

Given that both dQ/dT and dQ/dt are known, we have;

$\dfrac{dQ}{dT} = 0.18 \, (kcal/ ^{\circ} C)$

$\dfrac{dQ}{dt} = 12.9 \, (kcal/ min)$

Therefore, we get;

$\dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = {\dfrac{dQ}{dt} } \times \dfrac{dT}{dQ} = \dfrac{dT}{dt}$

$\dfrac{dT}{dt} = \dfrac{\dfrac{dQ}{dt} }{\dfrac{dQ}{dT} } = \dfrac{12.9 \, kcal / min }{0.18 \, kcal/ ^{\circ} C } = 71.\overline 6 \, ^{\circ } C/min$

For the sample, we have the change in temperature per minute, dT/dt, presented as follows;

$\dfrac{dT}{dt} = 71.\overline 6 \, ^{\circ } C/min$

8 0

3 years ago

BRO PLS I HAVE TO PASS THIS (click for question) 20 POINTS!

Masja [62]

Answer:

D, E.

Step-by-step explanation:

Simplify the expression inside the parentheses:

$3(23 + 18) < 6a$

$3(41) < 6a$

Multiply:

$123 < 6a$

Substitute each value provided for a:

Opt A. 123 < 6(18) = 123 < 108. Incorrect option.

Opt B. 123 < 6(19) = 123 < 114. Incorrect option.

Opt C. 123 < 6(20) = 123 < 120. Incorrect option.

Opt D. 123 < 6(22) = 123 < 132. Correct option.

Opt E. 123 < 6(24) = 123 < 144. Correct option.

3 0

3 years ago

The statistical difference between a process operating at a 5 sigma level and a process operating at a 6 sigma level is markedly

Svet_ta [14]

Answer:

True

Step-by-step explanation:

A six sigma level has a lower and upper specification limits between $\\ (\mu - 6\sigma)$ and $\\ (\mu + 6\sigma)$ . It means that the probability of finding no defects in a process is, considering 12 significant figures, for values symmetrically covered for standard deviations from the mean of a normal distribution:

$\\ p = F(\mu + 6\sigma) - F(\mu - 6\sigma) = 0.999999998027$

For those with defects operating at a 6 sigma level, the probability is:

$\\ 1 - p = 1 - 0.999999998027 = 0.000000001973$

Similarly, for finding no defects in a 5 sigma level, we have:

$\\ p = F(\mu + 5\sigma) - F(\mu - 5\sigma) = 0.999999426697$ .

The probability of defects is:

$\\ 1 - p = 1 - 0.999999426697 = 0.000000573303$

Well, the defects present in a six sigma level and a five sigma level are, respectively:

$\\ {6\sigma} = 0.000000001973 = 1.973 * 10^{-9} \approx \frac{2}{10^9} \approx \frac{2}{1000000000}$

$\\ {5\sigma} = 0.000000573303 = 5.73303 * 10^{-7} \approx \frac{6}{10^7} \approx \frac{6}{10000000}$

Then, comparing both fractions, we can confirm that a 6 sigma level is markedly different when it comes to the number of defects present:

$\\ {6\sigma} \approx \frac{2}{10^9}$ [1]

$\\ {5\sigma} \approx \frac{6}{10^7} = \frac{6}{10^7}*\frac{10^2}{10^2}=\frac{600}{10^9}$ [2]

Comparing [1] and [2], a six sigma process has 2 defects per billion opportunities, whereas a five sigma process has 600 defects per billion opportunities.

8 0

3 years ago

Please helpppp!?!!!!?

____ [38]

It would be 9!!!
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3 0

3 years ago

Read 2 more answers

Solve for y: 1 + 7y = 5x -2

Setler79 [48]

Answer:

5x/7 - 3/7