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katrin2010 [14]

3 years ago

10

The "Garbage Project" at the University of Arizona reports that the amount of paper discarded by households per week is normally

distributed with mean 9.4 lb and standard deviation 4.2 lb. What percentage of households throw out at least 9 lb of paper a week? (Round your answer to one decimal place.)

1 answer:

Lorico [155]3 years ago

7 0

Answer:

54%

Explanation:

We are given that

S.D=4.2 lb

Mean= $\mu=9.4 lb$

We have to find the percentage of household throw out at least 9 lb of paper a week.

Normal distribution formula :

$P(X\geq a)=P(z\geq \frac{a-\mu}{\sigma})$

We have a=9

$P(X\geq 9)=P(z\geq \frac{9-9.4}{4.2})=P(z\geq -0.1)$

$P(X\geq 9)=1-P(z$

$P(X\geq 9)=1-0.4602=0.5398\times 100=54$ %

Hence, the percentage of household throw out at least of paper a week=54%

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Answer:

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The full calculation can be found in the attachment below.

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One mole of iron (6 x 10^23 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-

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Answer:

Explanation:

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3 years ago

During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.148 kg baseball crashing through the pane

mr_godi [17]

Explanation:

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3 years ago

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balandron [24]

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6 0

3 years ago

Read 2 more answers

An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10

s2008m [1.1K]

Answer:

<em>1.228 x </em> $10^{-6}$ <em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x $10^{3}$ N

modulus of elasticity E = 85 GN/m^2 = 85 x $10^{9}$ Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = $\frac{\pi D^{2} }{4}$ = $\frac{3.142* 40^{2} }{4}$ = 1256.8 mm^2

area of hole a = $\frac{\pi(D^{2} - d^{2}) }{4}$ = $\frac{3.142*(40^{2} - 30^{2})}{4}$ = 549.85 mm^2

Total contraction of the bar = $\frac{F*L}{AE} + \frac{Fl}{aE}$

total contraction = $\frac{F}{E} * (\frac{L}{A} +\frac{l}{a})$

==> $\frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85})$ = <em>1.228 x </em> $10^{-6}$ <em> mm </em>

6 0

3 years ago