It’s a little complicated but here’s how it works:
Imagine a table with the intervals
0:4 , 4:6 , 6:7 , 7:10 , 10:13 (10 year intervals)
Then we have different rows
Class width: 4 , 2 , 1 , 3 , 3
Freq density: 0.2 , 0.5 , 1.2 , 0.7 , 0.3
So now calculate frequency where freq = class width * density
Freq: 0.8 , 1 , 3.6 , 2.1 , 0.9
So to find median find cumulative frequency
(Add all freq)
Cfreq = 8.4 now divide by 2 = 4.2
So find the interval where 4.2 lies.
0.8 + 1 = 1.8 + 3.6 = 5.6
So 4.2 (median) will lie in that interval 60-70 years.
Answer:
glide reflection
Step-by-step explanation:
Given that,
Total no of cards = 20
The ratio of animal cards to fruit cards is 2:3
No of animal cards = 20
Solution,
Let there are 2x animal cards and 3x fruit cards. ATQ,
2x+3x = 20
5x = 20
x = 4
Animal cards = 2x
= 2(4)
= 8
Fruit cards = 3x
= 3(4)
= 12
It is also mentioned that, of the cards, 1/3 have bananas on them. It means,
Hence, 4 cards have bananas on them.
Given:
The inequalities are:
a) 
b) 
To find:
The solution of inequalities by substituting the given values.
Solution:
a) We have,
After substituting the given values one by one, we get
False statement.
False statement.
True statement.
True statement.
Therefore the solutions of 
are 
.
b)
We have,
After substituting the given values one by one, we get
True statement.
False statement.
False statement.
False statement.
Therefore, the solution of 
is 
.
Answer:
D
Step-by-step explanation:
When dividing fractions, multiply the first number by the reciprocal of the second number
2/5 ÷ 1/3
First, find the reciprocal of the second number: 1/3
To find the reciprocal, flip the numerator (top number) and denominator (bottom number)
1/3-->3/1
Now, multiply 2/5 and 3/1
2/5 * 3/1
Multiply across the numerators and denominators
6/5
