Answer:
a

b

Step-by-step explanation:
From the question we are told that
The data is 87 91 86 82 72 91 60 77 80 79 83 96
Generally the point estimate for the mean is mathematically represented as

=> 
=>
Generally the point estimate for the standard deviation is mathematically represented as

=> 
=> 
Part a.
The domain is the set of x values such that
, basically x can be equal to -1/2 or it can be larger than -1/2. To get this answer, you solve
for x (subtract 1 from both sides; then divide both sides by 2). I set 2x+1 larger or equal to 0 because we want to avoid the stuff under the square root to be negative.
If you want the domain in interval notation, then it would be
which means the interval starts at -1/2 (including -1/2) and then it stops at infinity. So technically it never stops and goes on forever to the right.
-----------------------
Part b.
I'm going to use "sqrt" as shorthand for "square root"
f(x) = sqrt(2x+1)
f(10) = sqrt(2*10+1) ... every x replaced by 10
f(10) = sqrt(20+1)
f(10) = sqrt(21)
f(10) = 4.58257569 which is approximate
-----------------------
Part c.
f(x) = sqrt(2x+1)
f(x) = sqrt(2(x)+1)
f(x+2a) = sqrt(2(x+2a)+1) ... every x replaced by (x+2a)
f(x+2a) = sqrt(2x+4a+1) .... distribute
we can't simplify any further
5 : 25 → (÷5) → 1 : 5
Answer : 1 : 5
a) (2a - b)² = (4a² - 4ab + b²)
b) (10m - n²)² = (100m² - 20mn² + n⁴)
c) (4x - 4²) = (16x² - 8x + 4⁴)
d)
e)

f)

Answer:
B
Step-by-step explanation:
+ 6 = x ( subtract 6 from both sides )
= x - 6 ( square both sides )
x = (x - 6)² ← expand using FOIL
x = x² - 12x + 36 ( subtract x from both sides )
0 = x² - 13x + 36 , that is
x² - 13x + 36 = 0 ← in standard form
(x - 4)(x - 9) = 0 ← in factored form
Equate each factor to zero and solve for x
x - 4 = 0 ⇒ x = 4
x - 9 = 0 ⇒ x = 9
As a check
Substitute these values into the equation and if both sides are equal then they are the solutions.
x = 4
left side =
+ 6 = 2 + 6 = 8
right side = x = 4
Since 8 ≠ 4 then x = 4 is an extraneous solution
x = 9
left side =
+ 6 = 3 + 6 = 9
right side = x = 9
Thus the solution is x = 9 → B