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trasher [3.6K]
3 years ago
15

the amount of money spent weekly on cleaning maintenance and repairs at a large restaurant was observed over a long period of ti

me to be approximately normally distributed with a mean of $609 and standard deviation of $46
Mathematics
1 answer:
soldi70 [24.7K]3 years ago
4 0

Answer:


Step-Question: The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was ...

The amount of money spent weekly on cleaning, maintenance, and repairs at a large restaurant was observed over a long period of time to be approximately normally distributed, with mean ? = $615 and standard deviation ? = $38.

(a) If $646 is budgeted for next week, what is the probability that the actual costs will exceed the budgeted amount? (Round your answer to four decimal places.)

(b) How much should be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a given week is only0.08? (Round your answer to the nearest dollar.)

$by-step explanation:


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Let y(t) be the solution to y˙=3te−y satisfying y(0)=3 . (a) Use Euler's Method with time step h=0.2 to approximate y(0.2),y(0.4
OLEGan [10]

Answer:

  • y(0.2)=3, y(0.4)=3.005974448, y(0.6)=3.017852169, y(0.8)=3.035458382, and y(1.0)=3.058523645
  • The general solution is y=\ln \left(\frac{3t^2}{2}+e^3\right)
  • The error in the approximations to y(0.2), y(0.6), and y(1):

|y(0.2)-y_{1}|=0.002982771

|y(0.6)-y_{3}|=0.008677796

|y(1)-y_{5}|=0.013499859

Step-by-step explanation:

<em>Point a:</em>

The Euler's method states that:

y_{n+1}=y_n+h \cdot f \left(t_n, y_n \right) where t_{n+1}=t_n + h

We have that h=0.2, t_{0}=0, y_{0} =3, f(t,y)=3te^{-y}

  • We need to find y(0.2) for y'=3te^{-y}, when y(0)=3, h=0.2 using the Euler's method.

So you need to:

t_{1}=t_{0}+h=0+0.2=0.2

y\left(t_{1}\right)=y\left(0.2)=y_{1}=y_{0}+h \cdot f \left(t_{0}, y_{0} \right)=3+h \cdot f \left(0, 3 \right)=

=3 + 0.2 \cdot \left(0 \right)= 3

y(0.2)=3

  • We need to find y(0.4) for y'=3te^{-y}, when y(0)=3, h=0.2 using the Euler's method.

So you need to:

t_{2}=t_{1}+h=0.2+0.2=0.4

y\left(t_{2}\right)=y\left(0.4)=y_{2}=y_{1}+h \cdot f \left(t_{1}, y_{1} \right)=3+h \cdot f \left(0.2, 3 \right)=

=3 + 0.2 \cdot \left(0.02987224102)= 3.005974448

y(0.4)=3.005974448

The Euler's Method is detailed in the following table.

<em>Point b:</em>

To find the general solution of y'=3te^{-y} you need to:

Rewrite in the form of a first order separable ODE:

e^yy'\:=3t\\e^y\cdot \frac{dy}{dt} =3t\\e^y \:dy\:=3t\:dt

Integrate each side:

\int \:e^ydy=e^y+C

\int \:3t\:dt=\frac{3t^2}{2}+C

e^y+C=\frac{3t^2}{2}+C\\e^y=\frac{3t^2}{2}+C_{1}

We know the initial condition y(0) = 3, we are going to use it to find the value of C_{1}

e^3=\frac{3\left(0\right)^2}{2}+C_1\\C_1=e^3

So we have:

e^y=\frac{3t^2}{2}+e^3

Solving for <em>y</em> we get:

\ln \left(e^y\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y\ln \left(e\right)=\ln \left(\frac{3t^2}{2}+e^3\right)\\y=\ln \left(\frac{3t^2}{2}+e^3\right)

<em>Point c:</em>

To compute the error in the approximations y(0.2), y(0.6), and y(1) you need to:

Find the values y(0.2), y(0.6), and y(1) using y=\ln \left(\frac{3t^2}{2}+e^3\right)

y(0.2)=\ln \left(\frac{3(0.2)^2}{2}+e^3\right)=3.002982771

y(0.6)=\ln \left(\frac{3(0.6)^2}{2}+e^3\right)=3.026529965

y(1)=\ln \left(\frac{3(1)^2}{2}+e^3\right)=3.072023504

Next, where y_{1}, y_{3}, \:and \:y_{5} are from the table.

|y(0.2)-y_{1}|=|3.002982771-3|=0.002982771

|y(0.6)-y_{3}|=|3.026529965-3.017852169|=0.008677796

|y(1)-y_{5}|=|3.072023504-3.058523645|=0.013499859

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Darya [45]

Answer:

3x-4y=15

7x+y=4

Solve7x+y=4 for y:

7x+y=4

7x+y+−7x=4+−7x (Add -7x to both sides)

y=−7x+4

Step: Substitute −7x+4 for y in 3x−4y=15:

3x−4y=15

3x−4(−7x+4)=15

31x−16=15 (Simplify both sides of the equation)

31x−16+16=15+16 (Add 16 to both sides)

31x=31

(Divide both sides by 31)

x=1

Step: Substitute 1 for x in y=−7x+4:

y=−7x+4

y=(−7)(1)+4

y=−3(Simplify both sides of the equation)

Answer: x=1 and y=−3

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What is the solution to the equation 9^(x - 2) = 27 ?
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9^{x-2}=27\\\\(3^2)^{x-2}=3^3\\\\3^{2(x-2)}=3^3\\\\3^{2x-4}=3^3\iff2x-4=3\ \ \ |add\ 4\ to\ both\ sides\\\\2x=7\ \ \ \ |divide\ both\ sides\ by\ 2\\\\\boxed{x=3.5}
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