Find the area of this circle. Use 3 for pi.
A = pi r2
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3 ft
[?] ft2

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.

answer

a) the midrange of the data set shown is 85.5

step by step explanation

let's find the midrange.

the midrange is nothing but add the greatest value and the least value, then divide it by 2.

here the greatest value = 99

the least value = 72

midrange = (99 + 72)/2

= 171/2

midrange = 85.5

you.

the formula for converting absorbance to transmittance goes as:

a = log (1/t)  a = 2 - log(%t)  10^a = 1/t  t = 10^-a  %t = 100(10^-a)

so at the given value of a,   a = 1.5

%t = 100(10^-1.5)

%t = 3.16%

hey i have you need to go to the lesson of the dba and go to page 4 there should be cards and they have questions on them and if you click them there will be the answers and you can use that for the dba

step-by-step explanation: