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Sav [38]
3 years ago
9

Solve the quadratic equation x2 - x = 30 by completing the square

Mathematics
1 answer:
kap26 [50]3 years ago
5 0

Answer:

A

Step-by-step explanation:

Given

x² - x = 30

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- \frac{1}{2})x + \frac{1}{4} = 30 + \frac{1}{4}

(x - \frac{1}{2})² = \frac{121}{4}

Take the square root of both sides

x - \frac{1}{2} = ± \sqrt{\frac{121}{4} } = ± \frac{11}{2}

Add \frac{1}{2} to both sides

x = \frac{1}{2} ± \frac{11}{2}, thus

x = \frac{1}{2} - \frac{11}{2} = - 5

x = \frac{1}{2} + \frac{11}{2} = 6

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Anyone know this?????
kifflom [539]

Answer:

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Step-by-step explanation:

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6 0
3 years ago
A quality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to th
Burka [1]

Answer:

Confidence interval : 21.506 to 24.493

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

X=21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Sample mean = \bar{x}=\frac{\sum x}{n}

Sample mean = \bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}

Sample mean = \bar{x}=23

Sample standard deviation = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}

Sample standard deviation = \sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}

Sample standard deviation= s = 1.72

Degree of freedom = n-1 = 20-1 -19

Critical value of t using the t-distribution table t_{\frac{\alpha}{2} = 3.883

Formula of confidence interval : \bar{x} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}

Substitute the values in the formula

Confidence interval : 23 \pm 1.73 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 23 -3.883 \times \frac{1.72}{\sqrt{20}} to 23 + 3.883 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 21.506 to 24.493

Hence Confidence interval : 21.506 to 24.493

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( - 0.625, - 1.75 )

so since the first equation is already simplified , just replace it in the second equation

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