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3 years ago

If xy = 1, which is equivalent to x(x – 1)(y + 1)?

Mathematics

1 answer:

dmitriy555 [2]3 years ago

7 0

Answer:

B) x^2 – 1

Step-by-step explanation:

x(x – 1)(y + 1)

FOIL (x-1)(y+1)

firsy: xy

outer 1x

inner -1y

last (1)(-1) =-1

Add them together xy +x-y-1

x( xy +x-y-1)

Replace xy with 1

x( 1+x-y-1)

x(+x-y)

Distribute

x^2-xy

But xy=1

x^2 -1

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3 years ago

Read 2 more answers

Anne creates necklaces and rings to sell. It takes her 3 hours to make a necklace and 1 hour to make a ring. If she makes 10 ite

ankoles [38]

Hi! to solve the question, how many necklaces did anne make in 16 hours if she makes 10 items in 2 standard 8-hour work days, we should first check the time she needs to make each item.

3 individual hours to make a necklace, and 1 hour to make a ring. according to the problem, she made 10 items in 16 hours. this problem will most likely be trial and error. let’s try one first guess.

my first guess will be, maybe she will make 6 rings and 4 necklaces. 6 times 1 is 6, which takes 6 hours for the rings. 4 times 3 is equal to 12. 6 + 12 is too high! let’s try again.

what about 6 rings and 2 necklaces? 6 times 1 is 6, so that’s 6 hours. 2 times 3 is 6 hours! 6 plus 6 is 12, which is close, but not quite there!

let’s try 7 rings and 3 necklaces! 7 times 1 is 7 hours. 3 times 3 is equal to 9 hours! total that together, 7 plus 9, is equal to 16 hours spent!

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3 years ago

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 199 times. Here are the observed

Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28, 29, 47, 40, 22, 33.

Probability of an Event

$=\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}$

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

$(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12$

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

$(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911$