Answer:
BCD
Step-by-step explanation:
The top rectangle is ABCD because of its vertices. BCD is included in it.
Answer:
13/100 + 11/10 = 123/100
Step-by-step explanation:
Answer:
B) 81π units²
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Geometry</u>
Radius of a Circle Formula: r = d/2
Area of a Circle Formula: A = πr²
Step-by-step explanation:
<u>Step 1: Define</u>
Diameter <em>d</em> = 18 units
<u>Step 2: Manipulate Variables</u>
Radius <em>r</em> = 18 units/2 = 9 units
<u>Step 3: Find Area</u>
- Substitute in <em>r</em> [Area of a Circle Formula]: A = π(9 units)²
- [Area] Evaluate exponents: A = π(81 units²)
- [Area] Multiply: A = 81π units²
Answer:
Point estimate for the population variance = 3.92 *
.
Step-by-step explanation:
We are given that a sample of 5 strings of thread is randomly selected and the following thicknesses are measured in millimeters ;
X X -
1.13 1.13 - 1.188 = -0.058 3.364 * 
1.15 1.15 - 1.188 = -0.038 1.444 *
1.15 1.15 - 1.188 = -0.038 1.444 * 
1.24 1.24 - 1.188 = 0.052 2.704 * 
1.27 1.27 - 1.188 = 0.082 <u> 6.724 * </u>
<u> </u>
<u>= 0.01568 </u>
Firstly, Mean of above data,
=
=
= 1.188
Point estimate of Population Variance = Sample variance
=
=
= 3.92 *
.
Therefore, point estimate for the population variance = 3.92 *
.
Answer:
Height of the fighter plane =1.5km=1500 m
Speed of the fighter plane, v=720km/h=200 m/s
Let be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.
Muzzle velocity of the gun, u=600 m/s
Time taken by the shell to hit the plane =t
Horizontal distance travelled by the shell =u
x
t
Distance travelled by the plane =vt
The shell hits the plane. Hence, these two distances must be equal.
u
x
t=vt
u Sin θ=v
Sin θ=v/u
=200/600=1/3=0.33
θ=Sin
−1
(0.33)=19.50
In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell for any angle of launch.
H
max
=u
2
sin
2
(90−θ)/2g=600
2
/(2×10)=16km