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morpeh [17]
1 year ago
6

Let’s pretend that we want to give our employees in total a28% raise in the next two years, but we want to spread out aconsisten

t percent raise over two years (the same percent raise twice)to make this happen. Some justification for this could be: we can'tafford a 28% raise right now, but we don't want to lose ouremployees to the competition, so we want to start the raise processright away.a. What consistent percent raise should we give our employees each year to make our"in total a28% raise in the next two years" happen? (Be accurate with ananswer like 1.23%
Mathematics
1 answer:
Roman55 [17]1 year ago
5 0

Solution

- The solution steps are given below:

\begin{gathered} \text{ Let the original amount be X} \\  \\ \text{ If we want a 28\% raise over the next two years, then it means that the amount will be:} \\ X+\frac{28}{100}X=1.28X \\  \\ \text{ If we want to give them a consistent raise each year, we can compute the scenario as follows:} \\ \text{ Let the percentage be }y \\  \\ X+y\%\text{ of }X=\text{ Salary after first year} \\  \\ (X+y\%\text{ of }X)+y\%\text{ of }(X+y\%\text{ of }X)\text{ = Salary after second year.} \\  \\ \text{ But we already know that the salary after second year is }1.28X \\ \text{ Thus, we can say:} \\ (X+y\%\text{ of }X)+y\%\text{ of }(X+y\%\text{ of }X)=1.28X \\  \\ \text{ Simplifying, we have:} \\ X+\frac{yX}{100}+\frac{y}{100}(X+\frac{yX}{100})=1.28X \\  \\ \text{ Divide through by }X \\ 1+\frac{y}{100}+\frac{y}{100}(1+\frac{y}{100})=1.28 \\  \\ \text{ Subtract 1 from both sides and expand the brackets} \\ \frac{y}{100}+\frac{y}{100}+(\frac{y}{100})^2=1.28-1=0.28 \\  \\ \frac{2y}{100}+(\frac{y}{100})^2=0.28 \\  \\ \text{ Multiply both sides by }100^2 \\ 200y+y^2=2800 \\  \\ \text{ Rewrite, we have:} \\ y^2+200y-2800=0 \\  \\ Solving\text{ the equation using the Quadratic formula, we have that:} \\ y=\frac{-200\pm\sqrt{200^2-4(-2800)}(1)}{2(1)} \\  \\ y=-213.137\text{   or   }13.137 \\  \\ \text{ Since the change in salary is an increase, thus, the rate }y\text{ has to be positive.} \\ \text{ Thus, } \\ y=13.137\approx13.14\% \\  \\  \end{gathered}

Final Answer

y = 13.14%

The screenshots of the solution are:

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Answer:

Step-by-step explanation:

Part 1:

Let 

     Q₁ = Amount of the drug in the body after the first dose.

     Q₂ =  250 mg

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For Q₂,

we get:

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Therefore, after the second dose, 260 mg of the drug is present in the body.

Now, for Q₃ :

We get;

          Q₃ = 4% of Q2 + 250

               = 0.04 × 260 + 250

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For Q₄,

We get;

          Q₄ = 4% of Q₃ + 250 

                = 0.04 × 260.4 + 250

                = 10.416 + 250 

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Part 2:

To find out how large that amount is, we have to find Q₄₀.

Using the similar pattern

for Q₄₀,

We get;

           Q₄₀ = 250 + 250 × (0.04)¹ + 250 × (0.04)² + 250 × (0.04)³⁹

Taking 250 as common;

           Q₄₀ = 250 (1 + 0.04 + 0.042 + ⋯ + 0.0439)

                 = 2501 − 0.04401 − 0.04

           Q₄₀ = 260.4167

Hence, The greatest amount of antibiotics in Susan’s body is 260.4167 mg.

Part 3:

From the previous 2 components of the matter, we all know that the best quantity of the antibiotic in Susan's body is regarding 260.4167 mg and it'll occur right once she has taken the last dose. However, we have a tendency to see that already once the fourth dose she had 260.416 mg of the drug in her system, that is simply insignificantly smaller. thus we will say that beginning on the second day of treatment, double every day there'll be regarding 260.416 mg of the antibiotic in her body. Over the course of the subsequent twelve hours {the quantity|the quantity|the number} of the drug can decrease to 4% of the most amount, that is 10.4166 mg. Then the cycle can repeat.

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Step-by-step explanation:

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Step-by-step explanation:

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P(X=2) = 0.424

b2) P(X>2) = 1 - P(X≤2) = 1 - [P(X=0) + P(X=1) + P(X=2)]

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Mean of X = 1.67

C2) Standard Deviation of X

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Standard Deviation of X = 0.65

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3 years ago
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