Answer:
270 in
Step-by-step explanation:
the least amount of fabric she would need is equal to the area of the mat
the mat is in the shape of a rectangle
Area of a rectangle = length x width
15 x 18 = 270 in²
Formula of square area
a = s²
thus, area of the first square could be written as s₁² and area of the second square could be written as s₂²
First, write an expression in the form of equation system
An expression for "The combined area of two squares is 45 square centimeters" is:
⇒ s₁² + s₂² = 45 <em>(first equation)</em>
An expression for "Each side of one square is twice as long as side of the other" is:
⇒ s₁ = 2s₂ <em>(second equation)</em>
Second, solve the equation system by substitution method.
⇒ We could substitute 2s₂ to s₁ (second equation) in the first equation in order to find the value of s₂
s₁² + s₂² = 45
(2s₂)² + s₂² = 45
(2²)(s₂)² + s₂² = 45
4s₂² + s₂² = 45
5s₂² = 45
s₂² = 45/5
s₂² = 9
s₂² = 3²
s₂ = 3
⇒ Substitute the value of s₂ to the second equation in order to find the value of s₁
s₁ = 2s₂
s₁ = 2(3)
s₁ = 6
The length of each side of the larger square is 6 centimeters
(2 / x-1)> (7 / x + 1)
For this case we can solve the problem graphically.
The solution will be the shaded region.
Note: see attached image.
Answer:
The solution is given by:1 <x <9/5
Answer:
Step-by-step explanation:
Red:black = 1:1
Red cards = x
Black cards =x
x + x = 48
2x = 48
x = 48/2
x = 24
No. of red cards = 24
No of black cards = 24
Now 8 red cards are removed
So, No. of red cards = 24 - 8 =16
Red: Black = 16:24

Red: Black = 2:3
Answer: 0) 0.22
1) 0.20
2) 0.12
3) 0.14
4) 0.28
5) 0.04
<u>Step-by-step explanation:</u>
The total frequency is 33+30+18+21+42+6 = 150. This means they ran the experiment 150 times. The probability (P) is calculated by the satisfactory number of outcomes (frequency) divided by the total number of experiments/outcomes (total frequency):
![\begin{array}{c|c||lc}\underline{x}&\underline{f}&\underline{f\div 150}&\underline{\text{P}}\\0&33&33\div 150=&0.22\\1&30&30\div150=&0.20\\2&18&18\div 150=&0.12\\3&21&21\div 150=&0.14\\4&42&42\div 150=&0.28\\5&6&6\div 150=&0.04\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bc%7Cc%7C%7Clc%7D%5Cunderline%7Bx%7D%26%5Cunderline%7Bf%7D%26%5Cunderline%7Bf%5Cdiv%20150%7D%26%5Cunderline%7B%5Ctext%7BP%7D%7D%5C%5C0%2633%2633%5Cdiv%20150%3D%260.22%5C%5C1%2630%2630%5Cdiv150%3D%260.20%5C%5C2%2618%2618%5Cdiv%20150%3D%260.12%5C%5C3%2621%2621%5Cdiv%20150%3D%260.14%5C%5C4%2642%2642%5Cdiv%20150%3D%260.28%5C%5C5%266%266%5Cdiv%20150%3D%260.04%5Cend%7Barray%7D%5Cright%5D)