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3 years ago

Find exact solutions of

Mathematics

1 answer:

MrRa [10]3 years ago

5 0

Answer:

Step-by-step explanation:

cos(2x+√3)+cos 2x=0

2 cos (2x+√3+2x)/2 cos (2x+√3-2x)/2=0

or cos(2x+√3/2) cos √3/2=0

or cos (2x+√3/2)=0=cos (2n+1)π/2

or 2x+√3/2=(2n+1)π/2

or x=(2n+1)π/4+√3/4

x=π/4+√3/4=(π+√3)/4,(3π+√3)/4,(5π+√3)4,(7π+√3)/4,(9π+√3)/4,(11π+√3)/4,(13π+√3)/4,(15π+√3)/4

(4π=4×3.14=12.56 nearly)

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Answer:

P(2 quarters) = 19/ (13*16) = 19/208 = 0.0913

Step-by-step explanation:

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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

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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

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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 $23 + 3.883 \times \frac{1.72}{\sqrt{20}}$

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Hence Confidence interval : $21.506$ to $24.493$

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