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2 years ago

15

The concrete canoe team does some analysis on their design and calculates that they need a compressive strength of 860 psi. They

test several mix designs and chose one that has a sample mean of 900 psi with a standard error of the mean of 10 psi. The team wishes to be very conservative in their design and calculates a one-sided 99% confidence interval on the lower limit. What is this value?

1 answer:

vlada-n [284]2 years ago

8 0

Answer:

874 psi

Explanation:

Given a sample mean (x') = 900,

and a standard error (SE) = 10

At 99% confidence, Z(critical) = 2.58

That gives 99% confidence interval as,

x' ± Z(critical) x SE = 900 ± 2.58 x 10

The value of the lower limit is,

900 - 25.8 = 874.2

≈ 874 psi

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(a) Determine the dose (in mg/kg-day) for a bioaccumulative chemical with BCF = 103 that is found in water at a concentration of

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Answer:

0.064 mg/kg/day

6.25% from water, 93.75% from fish

Explanation:

Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.

The BCF = 10³, so the concentration of the chemical in the fish is:

10³ = x / (0.1 mg/kg)

x = 100 mg/kg

For 2 L of water and 30 g of fish:

2 kg × 0.1 mg/kg = 0.2 mg

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The total daily intake is 3.2 mg. Divided by the woman's mass of 50 kg, the dosage is:

(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day

b) The percent from the water is:

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And the percent from the fish is:

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3 years ago

Andreyy89

Answer:

7

Explanation:

5 + 2 = 7

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2 years ago

If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose. What happens to the speed of

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3 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

3 years ago