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2 years ago

72 kilograms equals how many pounds

1 answer:

4 0

1 kilogram = 2.20426 pounds

72 kilograms = ?? pounds

To find out how many pounds 72 kilograms equals to, just multiply 72 by 2.20462.

72* 2.20462 = 158.73264

72 kilograms = 158.73264 pounds

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188,888*1000 ???????<br><br><br><br> really need help!!!!!!!!!!!!!!!!!!!!!!!

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<h3>The answer is :-</h3>

188,888,000

<h3>Explanation :- </h3>

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so just add the zeros after 188,888

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1 year ago

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2 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 $24.493$

3 0

3 years ago

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A segment with endpoints A (2, 1) and C (4, 7) is partitioned by a point B such that AB and BC form a 3:2 ratio. Find B.

sdas [7]

$\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ A(2,1)\qquad C(4,7)\qquad \qquad \stackrel{\textit{ratio from A to C}}{3:2} \\\\\\ \cfrac{A\underline{B}}{\underline{B} C} = \cfrac{3}{2}\implies \cfrac{A}{C} = \cfrac{3}{2}\implies 2A=3C\implies 2(2,1)=3(4,7)\\\\[-0.35em] ~\dotfill\\\\ B=\left(\frac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \frac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\[-0.35em] ~\dotfill$

$\bf B=\left(\cfrac{(2\cdot 2)+(3\cdot 4)}{3+2}\quad ,\quad \cfrac{(2\cdot 1)+(3\cdot 7)}{3+2}\right)\implies B=\left( \cfrac{4+12}{5}~,~\cfrac{2+21}{5} \right) \\\\\\ B=\left(\cfrac{16}{5}~~,~~\cfrac{23}{5} \right)\implies B=\left( 3\frac{1}{5}~~,~~4\frac{3}{5} \right)$

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3 years ago

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