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3 years ago

Can someone help please

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

5 0

Hello there! I can help you! The formula for compound interest is P(1 + r)^t, where P= principal (initial amount), r = interest rate (in decimal form), and t = time (in years). Let's do this step by step. First off, we add the rate into 1. 4% is the interest rate (0.04 in decimal form). 1 + 0.04 is 1.04. Now, what we will do is raise that number to the 2nd power, because the time that elapses is 2 years. 1.04² is 1.0816. That's that. Now, multiply 7,500 to find the total amount of money. 1.0816 * 7,500 is 8,112. There. Toby's savings account balance in 2 years is £8,112.

Note: To solve for compound interest questions like it, add 1 to the percentage rate in decimal form, raise that number to a power based on the number of years (for example, raise the number to the 7th power if we are looking for the balance after 7 years), and then multiply that number by the starting amount. After you raise the number by a power, there may be a lot of numbers behind it. Whatever you do, DO NOT delete the number. Keep it there and multiply it by the principal.

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A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

$\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}$

(i) Set up an expression for the rate of change of salt concentration.

$\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}$

(ii) Integrate the expression

$\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C$

(iii) Find the constant of integration

$\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)$

(iv) Solve for A as a function of time.

$\text{The integrated rate expression is}\\\ln A = -0.003t + \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}$

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

$\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}$

b.Calculate the concentration

$A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}$

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is

$\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}$

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

$\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}$

(c) Concentration at infinite time

$\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}$

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

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Answer:

Step-by-step explanation:

A Parameter is a value that represents a property of a population. This could be the population mean.

A statistic is a value that represents a property of a sample. This could be the sample mean.

In the given scenario, the population is the entire 14 teams. This means that the percentage given is a value representing a property of the population. Therefore, the correct answer is

C. Parameter, because the data set of all 14 teams is a population.

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In a recently administered iq test, the scores were distributed normally, with mean 100 and standard deviation 15. what proporti

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To solve for proportion we make use of the z statistic. The procedure is to solve for the value of the z score and then locate for the proportion using the standard distribution tables. The formula for z score is:

z = (X – μ) / σ

where X is the sample value, μ is the mean value and σ is the standard deviation

when X = 70

z1 = (70 – 100) / 15 = -2

Using the standard distribution tables, proportion is P1 = 0.0228

when X = 130

z2 = (130 – 100) /15 = 2

Using the standard distribution tables, proportion is P2 = 0.9772

Therefore the proportion between X of 70 and 130 is:

P (70<X<130) = P2 – P1

P (70<X<130) = 0.9772 - 0.0228

P (70<X<130) = 0.9544

Therefore 0.9544 or 95.44% of the test takers scored between 70 and 130.

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