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3 years ago

Help me on this one please

Mathematics

1 answer:

shepuryov [24]3 years ago

3 0

Answer:

The equation is set up wrong

Step-by-step explanation:

$9^{2} +b^{2} =15^{2}$

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An important factor in selling a residential property is the number of times real estate, agents show a home. A sample of 21 hom

Natalka [10]

Using the t-distribution to build the 99% confidence interval, it is found that:

The margin of error is of 3.64.

The 99% confidence interval for the population mean is (19.36, 26.64).

<h3>What is a t-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm t\frac{s}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

t is the critical value.

n is the sample size.

s is the standard deviation for the sample.

The critical value, using a t-distribution calculator, for a two-tailed 95% confidence interval, with 21 - 1 = 20 df, is t = 2.086.

The other parameters are given as follows:

$\overline{x} = 23, s = 8, n = 21$

The margin of error is given by:

$M = t\frac{s}{\sqrt{n}} = 2.086\frac{8}{\sqrt{21}} = 3.64$

Hence the bounds of the interval are:

$\overline{x} - M = 23 - 3.64 = 19.36$

$\overline{x} + M = 23 + 3.64 = 26.64$

The 99% confidence interval for the population mean is (19.36, 26.64).

More can be learned about the t-distribution at brainly.com/question/16162795

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7 0

2 years ago

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

Alja [10]

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

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