2/3 is what percent of 1/4 ? of 1/6?

a grocer has 18 kg of black tea worth $3.5 per kg. she wants to mix it with green tea worth $2.6 per kg to sell the mixture at $2.9 per kg. how much of green tea should she use?

solve for both, and if you get it both correct, i will make you

it will be !

its rsm hw btw

Let...

x= kg of green tea

We can make an equation from the information given.

3.5(18)+2.6x=2.9(x+18)

63+2.6x=2.9x+52.2

10.8=0.3x

x=36

36 kg of green tea

I will also solve your first question.

2/3 is what percent of 1/4? 1/6?

Let's turn this question into an equation where we can solve for the unknown. Don't worry about the percent conversion yet.

2/3=1/4x

x=8/3

Now we can convert 8/3 to 800/3 %.

We solve the second part of the question similarly.

2/3=1/6x

x=12/3=4=400%

800/3% and 400%

1) If you want to know what percent is X in Y, you must do:

X/Y*100%

so in this case:

(2/3)/(1/4)*100% = 266.66%

So 2/3 is the 266.66% of 1/4

and

(2/3)/(1/6)*100% = 400%

So 2/3 is the 400% of 1/6

2) We have 18kg of black tea, and each kg cost $3.5

we want to mix it with X kg of green tea, (with a cost of $2.6) so the total cost per kg is $2.9

The final weight of the mixture will be 18kg + X

and the price of the each kilogram in the mixture will be:

P = (18kg*3.5 + X*%2.6)/(18kg + X) = $2.9

we must solve it for X

(18kg*3.5 + X*$2.6) = $2.9*(18kg + X) = $2.9*18kg + $2.9*X

X*($2.9 - $2.6) = ($3.5 - $2.9)*18kg

X = (0.6/0.3)*18kg = 36kg

So you need to add 36kg of green tea.

4th option is correct xy=-10

if this was to u plzz mark me the brainliest

the answer is: d) o, f, and kr.

atomic radius of bromine (br) is 114 pm.

atomic radius of oxygen (o) is 73 pm.

atomic radius of fluorine (f) is 72 pm.

atomic radius of krypron (kr) is 112 pm.

the atomic radius of a chemical element is a measure of the size of its atom.

the atomic radius varies with increasing atomic number, but usually increases because of increasing of number of electrons.

the atomic radius decreases across the periods because an increasing number of protons, because greater attraction between the protons and electrons.