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3 years ago

12

The National Aeronautics and Space Administration (NASA) purchased a new solar-powered battery guaranteed to have a failure rate

of only 1 in 20. A new system to be used in a space vehicle operates on one of these batteries. To increase the reliability of the system, NASA installed three batteries, each designed to operate if the other batteries in the chain fail. If the system is operated in a practical situation, what is the probability that all three batteries would fail

1 answer:

Reptile [31]3 years ago

5 0

Answer:

The probability that all three batteries would fail is given as:

= 0.000125

Step-by-step explanation:

Given that the probability of the new battery failing = 1/20 or 0.05,

Therefore, the probability that all three batteries would fail is given as:

= Probability of battery A failing * Probability of battery B failing * Probability of battery C failing

= P(A)*P(B)*P(C)

= 0.05 * 0.05 * 0.05

= 0.000125

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Step-by-step explanation:

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3 years ago

19) Albert says that the two systems of equations shown have the same solutions.

k0ka [10]

Answer:

option A) Agree, because the solutions are the same is correct.

Step-by-step explanation:

FIRST SYSTEM

$6x + y= 2$

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$\mathrm{Multiply\:}-x-y=-3\mathrm{\:by\:}6\:\mathrm{:}\:\quad \:-6x-6y=-18$

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adding the equation

$-6x-6y=-18$

$+$

$\underline{6x+y=2}$

$-5y=-16$

so the system becomes

$\begin{bmatrix}6x+y=2\\ -5y=-16\end{bmatrix}$

solve -5y for y

$-5y=-16$

Divide both sides by -5

$\frac{-5y}{-5}=\frac{-16}{-5}$

simplify

$y=\frac{16}{5}$

$\mathrm{For\:}6x+y=2\mathrm{\:plug\:in\:}y=\frac{16}{5}$

$6x+\frac{16}{5}=2$

subtract 16/5 from both sides

$6x+\frac{16}{5}-\frac{16}{5}=2-\frac{16}{5}$

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Divide both sides by 6

$\frac{6x}{6}=\frac{-\frac{6}{5}}{6}$

$x=-\frac{1}{5}$

Therefore, the solution to the FIRST SYSTEM is:

$x=-\frac{1}{5},\:y=\frac{16}{5}$

SECOND SYSTEM

$2x-3y = -10$

$-x-y=-3$

solving the system

$\begin{bmatrix}2x-3y=-10\\ -x-y=-3\end{bmatrix}$

$\mathrm{Multiply\:}-x-y=-3\mathrm{\:by\:}2\:\mathrm{:}\:\quad \:-2x-2y=-6$

$\begin{bmatrix}2x-3y=-10\\ -2x-2y=-6\end{bmatrix}$

$-2x-2y=-6$

$+$

$\underline{2x-3y=-10}$

$-5y=-16$

so the system of equations becomes

$\begin{bmatrix}2x-3y=-10\\ -5y=-16\end{bmatrix}$

solve -5y for y

$-5y=-16$

Divide both sides by -5

$\frac{-5y}{-5}=\frac{-16}{-5}$

Simplify

$y=\frac{16}{5}$

$\mathrm{For\:}2x-3y=-10\mathrm{\:plug\:in\:}y=\frac{16}{5}$

$2x-3\cdot \frac{16}{5}=-10$

$2x=-\frac{2}{5}$

Divide both sides by 2

$\frac{2x}{2}=\frac{-\frac{2}{5}}{2}$

Simplify

$x=-\frac{1}{5}$

Therefore, the solution to the SECOND SYSTEM is:

$x=-\frac{1}{5},\:y=\frac{16}{5}$

Conclusion:

As both systems of equations have the same solution.

Therefore, we conclude that Albert is right when says that the two systems of equations shown have the same solutions.

Hence, option A) Agree, because the solutions are the same is correct.

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