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3 years ago

X^2+2x-5=0 quadratic form

Mathematics

1 answer:

julsineya [31]3 years ago

3 0

Answer:

-1+√6

-1-√6

Step-by-step explanation:

x²+2x-5=0

x=(-2±√(2²-4(-5)))/2=(-2±√(4+20))/2=(-2±√24)/2

x₁=(-2+√24)/2 or x₂=(-2-√24)/2

x₁=(-2+2√6)/2 or (-2-2√6)/2

x₁=-1+√6 or x₂=-1-√6

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Select True or False.

nikdorinn [45]

Answer:

true

Step-by-step explanation:

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3 years ago

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For each level of precision, find the required sample size to estimate the mean starting salary for a new CPA with 95 percent co

Rzqust [24]

Answer:

(a) Margin of error ( E) = $2,000 , n = 54

(b) Margin of error ( E) = $1,000 , n = 216

(c) Margin of error ( E) = $500 , n= 864

Step-by-step explanation:

Given -

Standard deviation $\sigma$ = $7,500

$\alpha$ = 1 - confidence interval = 1 - .95 = .05

$Z_{\frac{\alpha}{2}}$ = $Z_{\frac{.05}{2}}$ = 1.96

let sample size is n

(a) Margin of error ( E) = $2,000

Margin of error ( E) = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

E = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$E^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{2000^{2}} \times 7500^{2}$

n = 54.0225

n = 54 ( approximately)

(b) Margin of error ( E) = $1,000

E = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

1000 = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$1000^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{1000^{2}} \times 7500^{2}$

n = 216

E = $Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}$

500 = $Z_{\frac{.05}{2}}\frac{7500}{\sqrt{n}}$

Squaring both side

$500^{2} = 1.96^{2}\times\frac{7500^{2}}{n}$

$n =\frac{1.96^{2}}{500^{2}} \times 7500^{2}$

n = 864

7 0

3 years ago

In 4 games, Larry averaged 23 PPG (points per game). How many points does he need to score in the last game to have an average o

ivann1987 [24]

Larry needs to score 33 points in the last game to have an average of 25 PPG.

To find the answer, we can solve the equation:

$\frac{(4)23+p}{5}=25$

(where p is how many points he needs to score in the last game)

To find the mean of a group of numbers, you add them all up and divide the total by the number of numbers there are. Since Larry averaged 23 PPG in 4 games, we can multiply 23 by 4 to get the total of the first 4 games from the data. Then, we find p, which we will add to get our final total. Then, you divide by the 5 games.

$\frac{(4)23+p}{5}=25$
$\frac{92+p}{5}=25$
${92+p}=125$
$p=33$

First, I simplified 4 x 23 to get 92. Then, I multiplied each side by 5 to get rid of the denominator. Finally, I subtracted 92 from each side to isolate p, and found that p = 33.

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3 years ago

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marta [7]

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3 years ago

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solong [7]

Answer:

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Step-by-step explanation:

divided by 3 to get the answer.

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2 years ago

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