Answer:
Tim has 64 dollars. I did the math and I know I'm right
Answer: Label the axis with numbers (x: from -10 to +10, y from 0 to 50)
Step-by-step explanation: Enter a dot for each coordinate.
Answer:
Step-by-step explanation:
we know that
The surface area of the square pyramid is equal to the area of the square base plus the area of its four triangular faces
so
![SA=b^2+4[\frac{1}{2}(b)(h)]](https://tex.z-dn.net/?f=SA%3Db%5E2%2B4%5B%5Cfrac%7B1%7D%7B2%7D%28b%29%28h%29%5D)
we have
substitute
![SA=1^2+4[\frac{1}{2}(1)(2)]](https://tex.z-dn.net/?f=SA%3D1%5E2%2B4%5B%5Cfrac%7B1%7D%7B2%7D%281%29%282%29%5D)
Answer:
The volume of water that remains on the cone is 523.6 cm³
Step-by-step explanation:
To solve this problem you have to keep in mind the formules that describes the volume of a cone and the volume of a sphere.
Volume of a cone = (πr²h)/3
Volume of a sphere = (4/3)πr³
So, if the base of the cone has a diameter of 10 cm, its radius is 5 cm. Its altitude is 10 cm. ⇒Volume = (πr²h)/3 ⇒ Volume = [π(5²)10) ⇒
Volume = 785.4 cm³. This is the initial volume of water.
Now if the sphere fits in the cone and half of it remains out of the water, the other half is inside the cone. Estimating the volume of the sphere and dividing it by two, you find the volume of water that was displaced.
Volume of a sphere = (4/3)πr³, here the radius is the same of the base of the cone (5 cm).
⇒ Volume = (4/3)π(5³) ⇒ Volume = 523.6 cm³ ⇒ The half of this volume is 261.8 cm³. This is the volume of water displaced.
⇒ The volume of water that remains on the cone is 523.6 cm³ (785.4 cm³- 261.8 cm³)
The following information about the mean and standard deviation has been provided:
ц=25, б=3,n=20
We need to compute Pr(X⁻≥26.3).
The corresponding z-value needed to be computed is:
Therefor, we get that
