First, set up your equation.
_(x+4)_ = 7
2
multiply both sides by 2 to cross out the 2 on the bottom
_(x+4)_ *2 = 7*2
2
x+4 =14
subtract 4 from both sides
x+4= 14
-4 -4
x=10
Answer:
The answer is x=20
Step-by-step explanation:
27-7=20
x=20
Answer:
Greatest sack = 42
1 candy bar and 3 lollipops
<em></em>
Step-by-step explanation:
Represent Candy bars with C and Lollipops with L
![C = 42](https://tex.z-dn.net/?f=C%20%3D%2042)
![L= 126](https://tex.z-dn.net/?f=L%3D%20126)
Solving (a): Greatest number of treat sacks
To solve this, we simply calculate the GCF of C and L
![42 = 2^1 * 3^1 * 7^1](https://tex.z-dn.net/?f=42%20%3D%202%5E1%20%2A%203%5E1%20%2A%207%5E1)
![126 = 2^1 * 3^2 * 7^1](https://tex.z-dn.net/?f=126%20%3D%202%5E1%20%2A%203%5E2%20%2A%207%5E1)
Hence, the GCF is
![GCF = 2^1 * 3^1 * 7^1](https://tex.z-dn.net/?f=GCF%20%3D%202%5E1%20%2A%203%5E1%20%2A%207%5E1)
![GCF = 42](https://tex.z-dn.net/?f=GCF%20%3D%2042)
Hence, greatest number of sack is 42
Solving (b): Number of treat in each sack.
To do this, we simply divide the number of C and L by the calculated GCF
For C:
![Treats = \frac{C}{GCF}](https://tex.z-dn.net/?f=Treats%20%3D%20%5Cfrac%7BC%7D%7BGCF%7D)
![Treats = \frac{42}{42}](https://tex.z-dn.net/?f=Treats%20%3D%20%5Cfrac%7B42%7D%7B42%7D)
![Treats = 1](https://tex.z-dn.net/?f=Treats%20%3D%201)
For L:
![Treats = \frac{L}{GCF}](https://tex.z-dn.net/?f=Treats%20%3D%20%5Cfrac%7BL%7D%7BGCF%7D)
![Treats = \frac{126}{42}](https://tex.z-dn.net/?f=Treats%20%3D%20%5Cfrac%7B126%7D%7B42%7D)
![Treats = 3](https://tex.z-dn.net/?f=Treats%20%3D%203)
<em>Hence, 1 candy bar and 3 lollipops</em>
Lumber was only about 36.95% of the bid.
Answer:
A fair six-sized die is rolled. Find the probability of getting at least a 4.
There are 6 outcomes and three of them are 4, 5 or 6, so the probability of greater than or equal to 4 is 3/6=½.