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4 years ago

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A 150-pound person burns 5.1 calories per minute when walking at a speed of 4 miles per hour. While walking, this person eats

a snack that has 60 calories. This snack subtracts from the calories burned while walking. How long must the person walk at this speed to burn at least 150 calories? Use pencil and paper. Describe what values the person could change so that the amount of time spent walking would be less.

1 answer:

Fantom [35]4 years ago

5 0

Answer:

He needs to walk 41 mins or 2,76 miles to burn 150 calories.

If he increases the speed of walking or eats snack with less calories, he will spend less time for walking.

Step-by-step explanation:

The person is burning

$5,1*60=306$

calories per hour.

He needs to burn 150 calories plus 60 calories that comes from the snack. In total 210 calories to burn.

$210/306=0,69$

He needs to walk 0,69 hours (aprrox. 41 mins) to burn 210 calories

$0,69*4=2,76$

In total he need to walk 2,76 miles to burn 210 calories.

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Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

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+ $\frac{x^2}{2!}$

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3 years ago

How many different committees can be formed from 11

Vikki [24]

Answer:

2 Committees

Step-by-step explanation:

2 Committees and 3/4 of a committee

There are only 11 teachers so that means only 2 committees can be filled with 4 teachers if that is the max amount per committee.

There are more than enough students so no need to worry on them.

11 divided by 4 = 2.75

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4 years ago

Find the perimeter of the triangle.

PtichkaEL [24]

Answer:

Perimeter of the Triangle

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3 years ago

If a student was randomly guessing while completing a 20 question multiple choice exam, what is the probability they would score

mario62 [17]

Answer:

you want 4 correct and 16 incorrect

there are 20 questions

each question has four answers, so

P(right answer) = 1/4

P(wrong answer) = 3/4

----

Since you want 4 correct of 20 we have a combination of 20C4

This is a binomial problem where p = 1/4, q = 3/4 and we get

(20 "choose" 4)*(probability correct)^(number correct)*(probability incorrect)^(number incorrect)

putting numbers in we get

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This gives us

~ .189685

Step-by-step explanation:

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4 years ago

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DIA [1.3K]

End of First Year (750/100)x12(3/5)=54 $750+54=804
End of Second Year (804/100)x12(3/5) = 58 $804+58=862
End Of Third Year (862/100)x12(3/5) = 62 $862+62=924

STOCK WORTH NOW = 924

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3 years ago