Answer: 1.5
Step-by-step explanation: you have to divide 6 by 4 and you get 1.5
Answer:
A reasonable way to use them is to code the five most-common strings in the ... bytes, compared to a bitmap size of only 6 × 8 bits = 6 bytes. ... Columns 2, 4, 5, and 7 yield the six runs 0,5,1,1,1,eol each. ... 2.3: Replacing 10 by 3 we get x = k log2 3 ≈ 1.58k. ... Table Ans.3: Probabilities and Entropies of Two Symbols.
Step-by-step explanation:
?
Let x = no. of 10 oz cups sold
Let y = no. of 14 oz cups sold
Let z = no. of 20 oz cups sold
:
Equation 1: total number of cups sold:
x + y + z = 24
:
Equation 2: amt of coffee consumed:
10x + 14y + 20z = 384
:
Equation 3: total revenue from cups sold
.95x + 1.15y + 1.50z = 30.60
:
Mult the 1st equation by 20 and subtract the 2nd equation from it:
20x + 20y + 20z = 480
10x + 14y + 20z = 384
------------------------ subtracting eliminates z
10x + 6y = 96; (eq 4)
Mult the 1st equation by 1.5 and subtract the 3rd equation from it:
1.5x + 1.5y + 1.5z = 36.00
.95x + 1.15y+ 1.5z = 30.60
---------------------------subtracting eliminates z again
.55x + .35y = 5.40; (eq 5)
Multiply eq 4 by .055 and subtract from eq 5:
.55x + .35y = 5.40
.55x + .33y = 5.28
--------------------eliminates x
0x + .02y = .12
y = .12/.02
y = 6 ea 14 oz cups sold
Substitute 6 for y for in eq 4
10x + 6(6) = 96
10x = 96 - 36
x = 60/10
x = 6 ea 10 oz cups
That would leave 12 ea 20 oz cups (24 - 6 - 6 = 12)
Check our solutions in eq 2:
10(6) + 14(6) + 20(12) =
60 + 84 + 240 = 384 oz
A lot steps, hope it made some sense! I hope this helps!! ;D
Answer:
You will have 48 blue streamers.
Step-by-step explanation:
This question can be solved by proportions, using rule of three.
We have that:
For 12 blue streamers, there are have 9 gold streamers. How many blue streamers are there when there are 36 gold streamers?
12 blue - 9 gold
x blue - 36 gold
Applying cross multiplication

Simplifying by 9

You will have 48 blue streamers.
Answer:
C
Step-by-step explanation:
We can easily rule out B and D because they have dotted lines and both equations are greater than or equal to and less than or equal to.