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3 years ago

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A rectangle has a length of 10 inches and a width of 5 inches. If the length is increased by x inches and the width decreased by

twice that amount, express the area of the rectangle as a function of x.

1 answer:

lorasvet [3.4K]3 years ago

5 0

Answer:12.5

Step-by-step explanation

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The difference between 15 and the product of 5 and y

eduard

The difference (-) between 15 and the product (x) of 15 and y

15y - 15

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3 years ago

A 150 pound woman would recieve 24 grams of a certain medicine. How many grams should a 120 pound woman receive?

eduard

Answer:

This is matter of ratio!

42/150 = x/120. It's 33.6 grams :D

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3 years ago

Rename 28 percent as a decimal and fraction in its simplest form.

VARVARA [1.3K]

28% is 28/100 as a fraction but the simplest form is 14/50 and it can be simplified to 7/25. the decimal is 0.28

fraction is 7/25
decimal is 0.28

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3 years ago

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Find the equation of the line that passes through (3,-4) and is parallel to 3x+y+2=0

miskamm [114]

Answer:

y=-3x+5

Step-by-step explanation:

it parallel to 3x+y+2=0 => line is: 3x+y+c=0

It through (3;-4) => 3(3)+(-4)+c=0 <=> c=-5

Line is 3x+y-5=0 => y=-3x+5

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3 years ago

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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 $\mu=24\cdot 0.018=0.432$

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

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3 years ago