Answer:
) Compute the sample mean x and the sample standard deviation s.
Do NOT use a computer. You may only use +, –, ×, ÷, and on a calculator.
Show ALL work.
9
143.1
9
15.5 16.2 16.1 15.8 15.6 16.0 15.8 15.9 16.2 = = = ∑ + + + + + + + +
n
x
x
i
= 15.9.
x x 2 x x − x ( ) 2
x − x
15.5
16.2
16.1
15.8
15.6
16.0
15.8
15.9
16.2
240.25
262.44
259.21
249.64
243.36
256.00
249.64
252.81
262.44
OR
15.5
16.2
Step-by-step explanation:
Answer: the answer is Triangles BCP and CDP are condruent
Step-by-step explanation:
Answer:
5.36
Step-by-step explanation:
Given that:
<BAD = <CAE, therefore, BD = EC
Let's take x to be the length of BD = EC
BD + DE + EC = BC
BC = 20,
BD = EC = x
DE ≈ 9.28
Thus,
x + 9.28 + x = 20
x + x + 9.28 = 20
2x + 9.28 = 20
Subtract 9.28 from both sides
2x + 9.28 - 9.28 = 20 - 9.28
2x = 10.72
Divided both sides by 2 to solve for x
BD ≈ 5.36
Answer:
x=8
Step-by-step explanation:
2(x+4) - 10 =14
distribute the 2 2x +8 -10 +14
simplify 2x-2=14
add 2 2x=16
devide by 2 x=8
Simple :)
Area of shaded part = are of 1/4 circle - area of both triangles.
Are of circle = pie r^2 so 100x3.14 = 314 cm2.
Area of triangle AOB= area of triangleDOE = bh/2= 5x10/2= 25 each
However, the traingles share a common area which is quad DOB(I)
Lets take traingle AOE, whose are is bh/2=10x10/2=50cm2.
50-area of triangle A(I)E= 50-(
