Answer:
ujn uzun g ne bu bu http TV fy ve g CD ve de c TT vcxcxc TR TRT TT ttgrsxfgzfggfxvzgrstgygfdfdzvsdcvzdcgz TR
Answer:
45
Step-by-step explanation:
Divide 30 by 10. You get 3. Take three and multiply it by 15. Your answer is 45.
Answer:
I use equal signs throughout but they are meant to be less than or equal to signs.
s=sandwiches
4.5s+6 less than or equal to 50
The answer is s is less than or equal to 12
Step-by-step explanation:
Finding Inequality
1. we know everything costs 50 in total so that is on one side of the equal sign.
____=50
2. we know he already spent 6 dollars so the money he can spend is 6 dollars less.
____+6=50
3. Each sandwich costs 4.50 $ so how much money does he have left to spend? we figure that out by multiplying the price by the number of sandwiches.
4.5s+6=50
Solving
1. 4.5s-6=50
First use the addition property of equality and add 6 to both sides.
2. 4.5s=56
Then use the division property of equality and divide each side by 4.5
3. you get a decimal but since you can't buy half a sandwich the answer is
s is less than or equal to 12
There are three unknowns in this problem. First, let's assign variables for these unknowns.
Let
x be the amount received by the first friend
y be the amount received by the second friend
z be the amount received by the third friend
Next, we formulate equations for the relationships
x = 4z
z = 4y
x + y + z = <span>$231,000
Solving simultaneously,
4(4y) + y + 4y = </span><span>$231,000
21y = </span><span>$231,000
y = $11,000
z = 4($11,000) = $44,000
x = 4($44,000) = $176,000</span>
Answer: 90.82%
Step-by-step explanation:
Given : The distribution of the amount of a certain brand of soda in 16 OZ bottles is approximately normal .
Mean : 
Standard deviation: 
Let X be the random variable that represents the amount of soda in bottles.
Formula for z-score : 
Z-score for 16 oz: 
Using the standard normal z-distribution table , the probability that the soda bottles that contain more than the 16 OZ is given by :_
Hence, the percentage of the soda bottles that contain more than the 16 OZ advertised is 90.82% .
