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3 years ago

How many cups does it take to make a gallon of water

Mathematics

1 answer:

Wittaler [7]3 years ago

8 0

It takes 16 cups of water to make a gallon

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Researchers are studying the distribution of subscribers to a certain streaming service in different populations. From a random

Katen [24]

Answer:

$CI = (0.17 - 0.27)\± 1.65\sqrt{\frac{(0.17)*(0.83) + (0.27)*(0.73)}{200}}$

Step-by-step explanation:

Given

$n = 200$

$x_1 = 34$ -- City C

$x_2 = 54$ --- City K

Required

Determine the 90% confidence interval

This is calculated using:

$CI = \bar x \± z\frac{\sigma}{\sqrt n}$

Calculating $\bar x$

$\bar x = \bar x_1 - \bar x_2$

$\bar x = \frac{x_1}{n} - \frac{x_2}{n}$

$\bar x = \frac{34}{200} - \frac{54}{200}$

$\bar x = 0.17 - 0.27$

For a 90% confidence level, the z-score is 1.65. So:

$z = 1.65$

Calculating the standard deviation $\sigma$

$\sigma = \sqrt{(\bar x_1)*(1 - \bar x_1) + (\bar x_2)*(1 - \bar x_2) }$

So:

$\sigma = \sqrt{(0.17)*(1 - 0.17) + (0.27)*(1 - 0.27) }$

$\sigma = \sqrt{(0.17)*(0.83) + (0.27)*(0.73)}$

So:

$CI = (0.17 - 0.27)\± 1.65\frac{\sqrt{(0.17)*(0.83) + (0.27)*(0.73)}}{\sqrt {200}}$

$CI = (0.17 - 0.27)\± 1.65\sqrt{\frac{(0.17)*(0.83) + (0.27)*(0.73)}{200}}$

7 0

2 years ago

Which expression is the result of factoring the expression below by taking out its greatest common factor?

Nat2105 [25]

Answer:

c is the correct answer

8 0

3 years ago

WILL GIVE BRAINLIEST!! ASAP <br> Which of the following expressions are equivalent to 6^-3

olga_2 [115]

Answer: 1/6^3 AND 1/216

Step-by-step explanation:

A negative number always goes to the denominator to become positive.

7 0

3 years ago

A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

3 0

3 years ago

22 must lie between the whole numbers<br> and

Crazy boy [7]

You should end of lab

3 0

2 years ago