Find the median of each set:-
Median is middle number of a data set. If a data set has an odd number of numbers then the median is the middle number when ordered form least to greatest but if its an even number you have to find the mean for the middle 2 numbers when ordered for least to greatest.
A.
1.2, 2.4, 3.2, 3.2, 3.6, 4.0, 4.1, 4.7
Even numbers = 8
3.2 + 3.6 = 6.8
6.8 ÷ 2 =
Median = 3.4
So this shows that A isn't the answer because the median of A is 3.4, not 3.2.
B.
1.6, 2.8, 2.9, 3.1, 3.3, 3.6, 4.2, 4.5
Even numbers = 8
3.1 + 3.3 = 6.4
6.4 ÷ 2 = 3.2
Median = 3.2
<span>So this shows that B is the answer because the median of B is 3.2.
C.
1.8, 2.0, 2.0, 2.2, 3.2, 4.7, 4.8, 4.9
</span>
Even numbers = 8
2.2 + 3.2 = 5.4
5.4 ÷ 2 = 2.7
Median = 2.7
<span>So this shows that C isn't the answer because the median of C is 2.7, not 3.2.
</span>
D.
1.4, 1.7, 2.9, 3.0, 3.1, 3.2, 3.2, 3.2, 4
Odd numbers = 9
Median = 3.1
<span>So this shows that D isn't the answer because the median of D is 3.1, not 3.2.
</span>
The stem and leaf plot which median is 3.2 is B.
Answer:
I don't know nan molla
Step-by-step explanation:
For example, Kino had a good opportunity to barely have a better life and he could have done anything to save his family but instead of doing that he refused to take the 1500 pesos.gddff
I would say the cube because it takes more space (volume) than the row of cubes
"Not sure if it makes sense but, eh...."
Answer:
The answer to your question is the big dog weighs 120 pounds
Step-by-step explanation:
Data
big dog = b
medium dog = m
small dog = s
Equations
b = 5s
m = 12 + s
s = 2/3 m
Process
1.- Substitute m in the last equation
s = 2/3(12 + s)
s = 24/3 + 2/3s
s - 2/3s = 8
1/3 s = 8
s = 8(3)
s = 24 pounds
2.- Substitute s in the first equation
b = 5(24)
b = 120 pounds
The answer is B) 4:5
First write the ratio out. To find the simplified expression, divide by the greatest common factor (GCF). In this case, the GCF of 28 and 35 is 7.
28:35
--- ----
7 7
=4:5
4:5 is the simplest form because you cannot simplify it further.
