What is the difference of the rational expressions below?

A phone company charges $0.45 for the first minute and $0.35 for each additional minute for a long distance call. Which numerica

Vinvika [58]

0.45+19(0.35)

Total cost $7.10 (0.45+6.65)

8 0

3 years ago

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What is the area of this figure? 49.5 in² 66.5 in² 84 in² 91 in²

Nookie1986 [14]

D 19 in2
find the area of a square, width times height
then find the area of the triangle, width times height divided by 2

6 0

3 years ago

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What is the answer to this question (3x3)+7-7=

exis [7]

Answer:

9

Step-by-step explanation:

Assuming x is the multiplication symbol, you would multiply 3 and the other 3 together. This leaves you with 9+7-7. Since 7-7=0 you can just get rid of both sevens completely. Now you are only left with 9.

6 0

3 years ago

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One division of a large defense contractor manufactures telecommunication equipment for the military. This division reports that

Zielflug [23.3K]

Answer:

At the 95% confidence level, management should not conclude that the percentage of rework for electrical components is lower than the rate of 12% for non-electrical components, since the upper bound of the confidene interval, which is 0.1339, is higher than 0.12.

Step-by-step explanation:

To reach a conclusion, we have to observe the confidence interval.

Should management conclude that the percentage of rework for electrical components is lower than the rate of 12% for non-electrical components?

Is the upper bound of the confidence interval lower than 12% = 0.12?

If yes, it should conclude that this percentage is lower.

Otherwise, it cannot conclude.

Confidence interval:

0.0758 to 0.1339

0.1339 is higher than 0.12.

So.

At the 95% confidence level, management should not conclude that the percentage of rework for electrical components is lower than the rate of 12% for non-electrical components, since the upper bound of the confidene interval, which is 0.1339, is higher than 0.12.

6 0

3 years ago

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Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$