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Anastaziya [24]
3 years ago
9

1 over 3b = 4 over 5. which of the following equals b in this equation?

Mathematics
1 answer:
Yuri [45]3 years ago
5 0

Answer:

b=2\frac{2}{5}

Step-by-step explanation:

we have

\frac{1}{3}b=\frac{4}{5}

solve for b

That means ---> isolate the variable b

Multiply by 3 both sides

\frac{1}{3}b(3)=\frac{4}{5}(3)

simplify

b=\frac{12}{5}

Convert to mixed number

\frac{12}{5}=\frac{10}{5}+\frac{2}{5}=2\frac{2}{5}

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4.9 x 10^4

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Dave’s Automatic Door, referred to in Exercise 29, installs automatic garage door openers. Based on a sample, following are the
HACTEHA [7]

The question is not complete and the full question says;

Calculate the (a) range, (b) arithmetic mean, (c) mean deviation, and (d) interpret the values. Dave’s Automatic Door installs automatic garage door openers. The following list indicates the number of minutes needed to install a sample of 10 door openers: 28, 32, 24, 46, 44, 40, 54, 38, 32, and 42.

Answer:

A) Range = 30 minutes

B) Mean = 38

C) Mean Deviation = 7.2

D) This is well written in the explanation.

Step-by-step explanation:

A) In statistics, Range = Largest value - Smallest value. From the question, the highest time is 54 minutes while the smallest time is 24 minutes.

Thus; Range = 54 - 24 = 30 minutes

B) In statistics,

Mean = Σx/n

Where n is the number of times occurring and Σx is the sum of all the times occurring

Thus,

Σx = 28 + 32 + 24 + 46 + 44 + 40 + 54 + 38 + 32 + 42 = 380

n = 10

Thus, Mean(x') = 380/10 = 38

C) Mean deviation is given as;

M.D = [Σ(x-x')]/n

Thus, Σ(x-x') = (28-38) + (32-38) + (24-38) + (46-38) + (44-38) + (40-38) + (54-38) + (38-38) + (32-38) + (42-38) = 72

So, M.D = 72/10 = 7.2

D) The range of the times is 30 minutes.

The average time required to open one door is 38 minutes.

The number of minutes the time deviates on average from the mean of 38 minutes is 7.2 minutes

4 0
3 years ago
Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.
Anika [276]

Answer:

\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \  }

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t

x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})

Now we can substitute in the equation

x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\

so the new equation is

y''(t)+ 8y'(t)-20y(t)=0

the auxiliary equation is

x^2+8x-20=0\\ x^2-2x+10x-20=0\\x(x-2)+10(x-2)=0\\(x+10)(x-2)=0\\ x=-10\text{ or }x=2

so the solutions of the new equation are

y(t)=ae^{2t}+be^{-10t}

with a and b real

as

x(t)=e^t\\ t(x)=ln(x)

y(x)=ae^{2ln(x)}+be^{-10ln(x)}=ax^2+bx^{-10}

hope this helps

do not hesitate if you have any questions

8 0
4 years ago
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