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The heaviest 16% of all men in weight weigh more than how many pounds ?

Likurg_2 [28]

Answer: 205 lbs

here are all the answers http://washburn.mpls.k12.mn.us/uploads/normal_calculations_practice_key.pdf

Step-by-step explanation:

5 0

3 years ago

Pls answer pls ok ok ok ok o ok

cestrela7 [59]

Answer: a

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Use the identity (x+y)(x2−xy+y2)=x3+y3 to find the sum of two numbers if the product of the numbers is 10, the sum of the square

professor190 [17]

Answer:

The sum of two numbers is 7.

Step-by-step explanation:

Given : the product of the numbers is 10, the sum of the squares of the numbers is 29, and the sum of the cubes of the numbers is 133.

Using identity $(x+y)(x^2-xy+y^2)=x^3+y^3$ and given details, we have to find the sum of two numbers.

Since, given that the product of the numbers is 10 that is $xy=10$

Also, given the sum of the squares of the numbers is 29 that is $x^2+y^2=29$

and the sum of the cubes of the numbers is 133 that is $x^3+y^3=133$

Using, the given identity $(x+y)(x^2-xy+y^2)=x^3+y^3$ ,

Substitute, the given values, we have,

$(x+y)(29-10)=133$

Simplify , we get,

$(x+y)(19)=133$

Divide both side by 19, we have,

$(x+y)=7$

Thus, the sum of two numbers is 7.

7 0

3 years ago

Read 2 more answers

i am doing bad in school i dont understand it i am feeling depressed wish someone would just do my work for me sigh

olga2289 [7]

Answer:

help you

Step-by-step explanation:

I will help you on somethings

6 0

3 years ago

Read 2 more answers

Y is directly proportional to square root of x<br> If y=56 when x=49 find,<br> y when x=81

STALIN [3.7K]

$\qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill } \\\\ \textit{\underline{x} varies directly with }\underline{z^5}\qquad \qquad \stackrel{\textit{constant of variation}}{x=\stackrel{\downarrow }{k}z^5~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}$

$\stackrel{\begin{array}{llll} \textit{"y" directly}\\ \textit{proportional to }\sqrt{x} \end{array}}{y = k\sqrt{x}}\qquad \textit{we know that} \begin{cases} y = 56\\ x = 49 \end{cases}\implies 56=k\sqrt{49} \\\\\\ 56=7k\implies \cfrac{56}{7}=k\implies 8=k~\hfill \boxed{y=8\sqrt{x}} \\\\\\ \textit{when x = 81, what is "y"?}\hfill y=8\sqrt{81}\implies y=8(9)\implies y=72$

7 0

2 years ago

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