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3 years ago

How do you complete the square

Mathematics

1 answer:

____ [38]3 years ago

8 0

divide all terms by x^2

move the number or term to the tight side of the equation

complete the square on the left side of the equation and balance them out by adding what you did on the right to the left

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Lydia is a childbirth assistant. The first year of her career, she attended the births of 2 babies. Every year after that, the n

Sonbull [250]

Answer:24 or 4

Step-by-step explanation:

2x2=4 4x6=24 24 or 4

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2 years ago

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Answer:

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3 years ago

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Suppose that you are provided with the following data for your country for a particular month: 200 million people are working, 2

lisov135 [29]

Answer:

$Unemployement\,\,rate=23%$

Step-by-step explanation:

To find the unemployment rate we have the next expression:

$Unemplyment\,\,rate=\frac{Unemployment\,\,people}{Total\,\, labor\,\, force}\cdot 100$

We have the next data:

200 million employed

20 million unemployed

40 million umemployed are not looking for a job

Then the total labor force will be the employed and the unemployed population then de have 260 million

$Unemployement\,\,rate=\frac{60million}{260million}\cdot 100$

$Unemployement\,\,rate=\frac{6}{26}\cdot 100$

$Unemployement\,\,rate=0.23\cdot 100$

$Unemployement\,\,rate=23%$

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3 years ago

Which of these numbers is greater than 2.25 but less than 1? Is it 1 or 1.5 or 0 or 2.5

Sloan [31]

Answer:

the answer is 1.5 because converting them into their fractional form indeed informs u that 1.5 is thus greater than 2.25 BT lesser than 1

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3 years ago

The long-distance calls made by the employees of a company are normally distributed with a mean of 6.3 minutes and a standard de

qaws [65]

Answer: a. 0.6759 b. 0.3752 c. 0.1480

Step-by-step explanation:

Given : The long-distance calls made by the employees of a company are normally distributed with a mean of 6.3 minutes and a standard deviation of 2.2 minutes

i.e. $\mu = 6.3$ minutes

$\sigma=2.2$ minutes

Let x be the long-distance call length.

a. The probability that a call lasts between 5 and 10 minutes will be :-

$P(5$

b. The probability that a call lasts more than 7 minutes. :

$P(X>7)=P(\dfrac{X-\mu}{\sigma}>\dfrac{7-6.3}{2.2})\\\\=P(Z>0.318)\ \ \ \ [z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Z$

c. The probability that a call lasts more than 4 minutes. :

$P(X$

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3 years ago