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gogolik [260]
3 years ago
8

-18 - 3 /4 v = 3 I need help with this who is smart enough to solve it

Mathematics
2 answers:
Blababa [14]3 years ago
8 0

Answer:

v =-28

Step-by-step explanation:

-18 - 3 /4 v = 3

Add 18 to each side

-18+18 - 3 /4 v = 3+18

-3/4 v = 21

Multiply each side by -4/3 to isolate v

-4/3 *-3/4v = 21*-4/3

v =-28

Pavel [41]3 years ago
7 0

Answer: v= -28

Step-by-step explanation:

Solve for v by simplifying both sides of the equation, then isolating the variable.

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The number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a m
Mazyrski [523]

Answer:

The probability that the instrument does not fail in an 8-hour shift is P(X=0) \approx 0.8659

The probability of at least 1 failure in a 24-hour day is P(X\geq 1 )\approx 0.3508

Step-by-step explanation:

The probability distribution of a Poisson random variable X representing the number of successes occurring in a given time interval or a specified region of space is given by the formula:

P(X)=\frac{e^{-\mu}\mu^x}{x!}

Let X be the number of failures of a testing instrument.

We know that the mean \mu = 0.018 failures per hour.

(a) To find the probability that the instrument does not fail in an 8-hour shift, you need to:

For an 8-hour shift, the mean is \mu=8\cdot 0.018=0.144

P(X=0)=\frac{e^{-0.144}0.144^0}{0!}\\\\P(X=0) \approx 0.8659

(b) To find the probability of at least 1 failure in a 24-hour day, you need to:

For a 24-hour day, the mean is \mu=24\cdot 0.018=0.432

P(X\geq 1 )=1-P(X=0)\\\\P(X\geq 1 )=1-\frac{e^{-0.432}0.432^0}{0!}\\\\P(X\geq 1 )\approx 0.3508

3 0
3 years ago
X+5 divided by 3=15 what is the answer?
GREYUIT [131]
All you have to do is solve for x. See the following steps below.

Step 1. Subtract \frac{5}{3} from both sides

X= 15-\frac{5}{3}

Step 2. Simplify 15- \frac{3}{5} to \frac{40}{3}

X= \frac{40}{3}
4 0
3 years ago
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Answer:

The runners drank 661 bottles of water.

Step-by-step explanation:

Subtract 661 from 750, and you are left with 89.

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Need a picture or some kind of information
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