We know that
red marbies=3------> r=3
green marbies=4----> g=4
blue marbies=3-----> b=3
total marbles in the bag=r+g+b-----> 3+4+3-----> 10
a) <span>probability of randomly selecting a green (g) marble
P(green)=g/total marbles------> 4/10-----> 2/5
b) </span>probability of randomly selecting a red (r) marble
the marble is replaced so we have the same total marbles
P(red)=r/total marbles------> 3/10
c) <span>Total probability of selecting this path is:
P(green)*P(red)-----> (2/5)*(3/10)-----> 6/50----> 3/25----> 0.12 (12%)
the answer is
0.12 (12%)</span>
Answer:
Part 1) 
Part 2) 
Step-by-step explanation:
we have
![f(x)=\sqrt[3]{x} +1](https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7Bx%7D%20%2B1)
Part 1) Find f(125)
we know that
f(125) is the value of the function f(x) when the value of x is equal to 125
so
For x=125
substitute in the function
![f(125)=\sqrt[3]{125} +1](https://tex.z-dn.net/?f=f%28125%29%3D%5Csqrt%5B3%5D%7B125%7D%20%2B1)
Remember that
substitute
![f(125)=\sqrt[3]{5^{3}} +1](https://tex.z-dn.net/?f=f%28125%29%3D%5Csqrt%5B3%5D%7B5%5E%7B3%7D%7D%20%2B1)
Applying the power of rule
![\sqrt[3]{5^{3}}=(5^{3})^{\frac{1}{3}}= (5)^{3*\frac{1}{3}}=5](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B5%5E%7B3%7D%7D%3D%285%5E%7B3%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D%20%285%29%5E%7B3%2A%5Cfrac%7B1%7D%7B3%7D%7D%3D5)
substitute
Part 2) Find f(-64)
we know that
f(-64) is the value of the function f(x) when the value of x is equal to -64
so
For x=-64
substitute in the function
![f(-64)=\sqrt[3]{-64} +1](https://tex.z-dn.net/?f=f%28-64%29%3D%5Csqrt%5B3%5D%7B-64%7D%20%2B1)
Remember that
substitute
![f(-64)=\sqrt[3]{-4^{3}} +1](https://tex.z-dn.net/?f=f%28-64%29%3D%5Csqrt%5B3%5D%7B-4%5E%7B3%7D%7D%20%2B1)
Applying the power of rule
![\sqrt[3]{-4^{3}}=(-4^{3})^{\frac{1}{3}}= (-4)^{3*\frac{1}{3}}=-4](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-4%5E%7B3%7D%7D%3D%28-4%5E%7B3%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%3D%20%28-4%29%5E%7B3%2A%5Cfrac%7B1%7D%7B3%7D%7D%3D-4)
substitute
Answer:
Explanation:
The area of the <em>octagon</em> may be calculated as the difference of the area of the original square and the area of the four corners cut off.
1) <u>Area of the square</u>.
The original square's side length is the same wide of the formed octagon: 10 cm.
So, the area of such square is: (10 cm)² = 100 cm².
2) <u>Area of the four corners cut off</u>.
Since, the corners were cut off two centimeters from each corner, the form of each piece is an isosceles right triangle with legs of 2 cm.
The area of each right triangle is half the product of the legs (because one leg is the base and the other leg is the height of the triangle).
Then, area of one right triangle: (1/2) × 2cm × 2cm = 2 cm².
Since, they are four pieces, the total cut off area is: 4 × 2 cm² = 8 cm².
3) <u>Area of the octagon</u>:
- Area of the square - area of the cut off triangles = 100 cm² - 8cm² = 92 cm².
And that is the answer: 92 cm².
16 * 4 + 7 * 8
64 + 56
120
Answer: 120 students
Answer:
6k +19
Step-by-step explanation:
7k-k+19
Combine like terms
7k-k = 6k
The expression becomes 6k +19
