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3 years ago

What is the distance between points S and U if S is 2, 1 and U is 6, 8

Mathematics

2 answers:

sleet_krkn [62]3 years ago

5 0

Answer:

Step-by-step explanation:

S has coordinates (2 , 1) and U has coordinates (6, 8) . Computing the SU vector gives us (4 , 7)

taking the distance formula : sqrt(4^2 +7^2)

= sqrt(65) = 8.1 which is answer C

balandron [24]3 years ago

4 0

The distance formula is:

D = sqrt((x2-x1)^2 + (y2-y1)^2)

D= sqrt ((6-2)^2 + (8-1)^2)

D = sqrt(4^2 + 7^2)

D = sqrt( 16 + 49)

D = sqrt(65)

D = 8.06

Rounded to the beat tenth = 8.1 units.

The answer is C. 8.1 units

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2 years ago

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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 $24.493$

3 0

3 years ago

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An athlete ran 800 meters in 160 seconds. What was his rate in meters per minute? Enter your answer in the box.

FromTheMoon [43]

The athlete's average speed was 300 meters/minute.

<h3>How to calculate the speed of the athlete?</h3>

To calculate the speed of the athlete we must perform the following operations.

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Transform the value from seconds to minutes, for which we must multiply 5 by 60 because each minute is made up of 60 seconds.

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According to the above, the average speed of the athlete is 300m/min.

Learn more about speed in: brainly.com/question/7359669

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