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3 years ago

Solve for n -112 = 7n

Mathematics

2 answers:

Eddi Din [679]3 years ago

8 0

Answer:

n -112 = 7n

Step-by-step explanation:

-6n= 112

-3n= 56

n= 56/ -3

n= -56/3

erik [133]3 years ago

6 0

Answer:

n= -56/3

Step-by-step explanation:

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Write each of the following expressions without using absolute value. |z−7|−|z−9|, if z<7

aivan3 [116]

Here's my thinking: If z<7, then z-7 is negative, and |z-7| is positive. So, with the understanding that z-7 is positive, the first part of this expression becomes:

7-z for z<7

Similarly: if z<7, then z-9 is negative and |z-9| is positive. Thus, -|z-9| is equivalent to (9-z).

|z−7|−|z−9|, if z<7 becomes (7-z) - (9-z) for z<7. Simplifying this, we get:

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4 years ago

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Harlamova29_29 [7]

Answer:

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Step-by-step explanation:

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3 years ago

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notsponge [240]

Answer:

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Step-by-step explanation:

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3 years ago

A system has a continuous load requirement of 80 MW. Find the expected load loss and the probability of load loss (loss of load

MariettaO [177]

Answer:

Expected load loss = 1.35

P (Load Loss) = 0.127

Step-by-step explanation:

The probability of failure is: q = 0.015

The probability of success is: p = 1 - q = 1 - 0.015 = 0.985

The number of units the generation system is composed of is, n = 9.

The capacity of each unit is 10 MW.

Let X = number of units that failed or are unavailable.

The random variable X follows a Binomial distribution with parameters n and q.

The probability function is:

$P (X=x)={n\choose x}p^{x}(1-p)^{n-x}$ ; x =0, 1, 2, 3...

The expected number of failures is:

$E(X)=nq=9\times0.015=0.135$

Then the expected load loss is:

$E (Load\ Loss)=10\times E(X)=10\times 0.135=1.35$

Thus, the expected load loss is 1.35.

The load loss will be when at least one unit fails.

The probability that at least one unit fails is:

P (X ≥ 1) = 1 - P (X < 1) = 1 - P (X = 0)

$=1-[{9\choose 0}(0.015)^{0}(1-0.015)^{9-0}]\\=1-0.8728\\=0.12718\\\approx0.127$

Thus, the probability of load loss is 0.127.

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Answer:

Step-by-step explanation:

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