Answer:
24 unit
Step-by-step explanation:
Given,
The vertices of the triangle ABC are,
A (1,1), B (7,1), and C (1,9),
By the distance formula,



Thus, the perimeter of the triangle ABC = AB + BC + CA = 6 + 10 + 8 = 24 unit
Try the last one; it doesn't mention anything foreign.
Let's say you want to compute the probability

where

converges in distribution to

, and

follows a normal distribution. The normal approximation (without the continuity correction) basically involves choosing

such that its mean and variance are the same as those for

.
Example: If

is binomially distributed with

and

, then

has mean

and variance

. So you can approximate a probability in terms of

with a probability in terms of

:

where

follows the standard normal distribution.
Part A)
The table doesn't represent y as a function of x.
Why? <span>One reason is that 1 or 3 is the first element in more than one ordered pair.
Part B)
f(x) = 10x + 20
f(120)= 10*120 + 20 = 1200 + 20 = 1220
</span><span>f(120) represents cost of boating for 120 hours.</span><span>
</span>
Answer:
y - 3y^2 + 4y^2 - 12y
y^2 - 11y
Step-by-step explanation: