Answer:
78.55m^2
Step-by-step explanation:
General formula for the area of a circle is πr^2
π=3.12
r=5m
So let's apply the formula
3.142×(5)^2
3.142×25
78.55
So the area is 78.55m^2
Answer:
1 st questions answer is third option
while the answer of other question is 3
Step-by-step explanation:
2nd qestions explanation
put value of x=2 in equation as x-2=0 from equation
(2)^3-6(2)^2+a2+10=2
8-24+a2+10=0
2a=-18+24
2a=6
a=3
Answer:
0.8 Not Outlier
1.1 Not Outlier
10.2 Not Outlier
10.9 Not Outlier
Solution:
Arranging the numbers in ascending order:
0.8 1.1 4.9 5.2 5.8 5.9 6.1 6.1 7.4 10.2 10.9
we can see that the median is 5.9.
We can find the first quartile Q1 by getting the median in the lower half of the data
0.8 1.1 4.9 5.2 5.8
that is, Q1 = 4.9
We can find the third quartile Q3 by getting the median for the upper half of the data
6.1 6.1 7.4 10.2 10.9
that is, Q3 = 7.4
We subtract Q1 from Q3 to find the interquartile range IQR.
IQR = Q3 - Q1 = 7.4 - 4.9 = 2.5
We can now calculate for the upper and lower limits:
upper limit = Q3 + 1.5*IQR = 7.4 + (1.5*2.5) = 11.15
lower limit = Q1 – 1.5*IQR = 0.8 - (1.5*2.5) = -2.95
There is no data point that lies above the upper limit and below the lower limit, therefore, there are no outliers in the data set.
For this all you have to do is plug in the x and y values into the equation and see if it is a true statement. If it's true then it's on the line. The x value is 2 and the y is -4.
12x + 3y = 6
12 (2) + 3 (-4) = 6
24 -12 = 6
12 = 6
The statement 12=6 is not a true statement so that point is not on the line.
