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rjkz [21]
2 years ago
13

A home building company routinely orders standard interior doors with a height of 80 inches. Recently the installers have compla

ined that the doors are not the standard height. The quality control inspector for the home building company is concerned that the manufacturer is supplying doors that are not 80 inches in height. In an effort to test this, the inspector is going to gather a sample of the recently received doors and measure the height. The alternative hypothesis for the statistical test to determine if the doors are not 80 inches is
Mathematics
1 answer:
Helga [31]2 years ago
3 0

Answer: Alternative Hypothesis

Ha : u ≠ 80 inches ( mean not equal to 80 inches)

Step-by-step explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean.

While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.

For the case above, the mean door height is 80 inches

H0 : u = 80 inches

Ha : u ≠ 80 inches ( mean not equal to 80 inches)

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Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
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Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q  = -t/100 + c

Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c}  = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}

when t = 0, Q = 200 L × 1 g/L = 200 g

Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

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Step-by-step explanation: true

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