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3 years ago

If there are 2.2 pounds in 1 kilogram how many pounds are there in x kilograms?

Mathematics

1 answer:

nirvana33 [79]3 years ago

5 0

Answer:

P=x/2.2

Step-by-step explanation:

We can create an equation to solve this.

Let the number of pounds be P

Let the number of kilograms be X

Now, we know that for 1 kilogram, there is 2.2 pounds.

Therefore if we have 1 kilogram (X) then we will have 2.2 pounds (P).

X=2.2P

All you have to do now is rearrange the equation to find the number of pounds in x kilograms. To do this divide both sides by 2.

Therefore you will get P=x/2.2

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PLS help its the last question

cluponka [151]

Answer:

Step-by-step explanation:

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2 years ago

A product is made up of components A, B, C, D, E, F, G, H, I, and J. Components A, B, C, and F have a 1/10,000 chance of failure

lesantik [10]

Answer:

The overall reliability is 99.7402 %

Step-by-step explanation:

The overall reliability of the product is calculated as the product of the working probability of the components.

For components A,B,C and F we have :

$P(failure)=\frac{1}{10000}$

⇒

$P(Work)=1-\frac{1}{10000}=\frac{9999}{10000}=0.9999$

For components D,E,G and H we have :

$P(failure)=\frac{3}{10000}$

⇒

$P(Work)=1-\frac{3}{10000}=\frac{9997}{10000}=0.9997$

Finally, for components I and J :

$P(failure)=\frac{5}{10000}$

⇒

$P(Work)=1-\frac{5}{10000}=\frac{9995}{10000}=0.9995$

Now we multiply all the working probabilities. We mustn't forget that we have got ten components in this case :

Components A,B,C and F with a working probability of 0.9999

Components D,E,G and H with a working probability of 0.9997

Components I and J with a working probability of 0.9995

Overall reliability = $(0.9999)^{4}(0.9997)^{4}(0.9995)^{2}=0.997402$

0.997402 = 99.7402 %

4 0

3 years ago

Find lim x→3 sqrt 2x+3-sqrt 3x/ x^2-3x. you must show your work or explain your work in words plsss I need help

sineoko [7]

I'm assuming the limit is supposed to be

$\displaystyle\lim_{x\to3}\frac{\sqrt{2x+3}-\sqrt{3x}}{x^2-3x}$

Multiply the numerator by its conjugate, and do the same with the denominator:

$\left(\sqrt{2x+3}-\sqrt{3x}\right)\left(\sqrt{2x+3}+\sqrt{3x}\right)=\left(\sqrt{2x+3}\right)^2-\left(\sqrt{3x}\right)^2=-(x-3)$

so that in the limit, we have

$\displaystyle\lim_{x\to3}\frac{-(x-3)}{(x^2-3x)\left(\sqrt{2x+3}+\sqrt{3x}\right)}$

Factorize the first term in the denominator as

$x^2-3x=x(x-3)$

The $x-3$ terms cancel, leaving you with

$\displaystyle\lim_{x\to3}\frac{-1}{x\left(\sqrt{2x+3}+\sqrt{3x}\right)}$

and the limand is continuous at $x=3$ , so we can substitute it to find the limit has a value of -1/18.

7 0

3 years ago

1. Use a piece of wool, thread or chord to measure the circumference of each of the circles found on the last page. To do this j

spayn [35]

Answer:

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Step-by-step explanation:

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3 years ago

Simplify. 7/8+(−2/3) divided by 5/6

Sedaia [141]

7/8+(−2/3) divided by 5/6

(7/8 - 2/3)/(5/6)
=(21/24 - 16/24) / (5/6)
= (5/24) / (5/6)
= 5/24 * 6/5
= 6/24
= 1/4

7 0

4 years ago

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