*see attachment for the figure referred to
Answer/Step-by-step explanation:
1. PN = 29
MN = 13
PM = ?
(Segment addition postulate)
(subtract MN from each side)
(substitute)
2. PN = 34, MN = 19, PM = ?
(sediment addition postulate)
(subtract MN from each side)
(substitute)
3. PM = 19, MN = 23, PN = ?
(Segment addition postulate)
(substitute)
4. MN = 82, PN = 105, PM = ?
(segment addition postulate)
(subtract MN from each side)
(substitute)
5. PM = 100, MN = 100, PN = ?
(Segment addition postulate)
(substitute)
Hello :
x : length y :<span> width
</span>xy = 50 ...(1)
x = 2y ...(2)
dubsct in (1) :
(2y)(y) = 50
y² = 25
y = 5
x = 10
Answer:
1/8
Step-by-step explanation:
yay
Exercise 1:
exponential decay:
The function is given by:
y = A (b) ^ ((1/3) * t)
Where,
A = 600
We look for b:
(480/600) * (100) = 80%
b = 0.8
Substituting:
y = 600 * (0.8) ^ ((1/3) * t)
We check for t = 6
y = 600 * (0.8) ^ ((1/3) * 6)
y = 384
Answer:
exponential decay:
y = 600 * (0.8) ^ ((1/3) * t)
Exercise 2:
linear:
The function is given by:
y = ax + b
Where,
a = -60 / 2 = -30
b = 400
Substituting we have:
y = -30 * x + 400
We check for x = 4
y = -30 * 4 + 400
y = 280
Answer:
linear:
y = -30 * x + 400
Exercise 3:
exponential growth:
The function is given by:
y = A (b) ^ ((1/3) * t)
Where,
A = 512
We look for b:
(768/512) * (100) = 150%
b = 1.5
Substituting:
y = 512 * (1.5) ^ ((1/2) * t)
We check for t = 4
y = 512 * (1.5) ^ ((1/2) * 4)
y = 1152
Answer:
exponential growth:
y = 512 * (1.5) ^ ((1/2) * t)
Answer:
a)0.08 , b)0.4 , C) i)0.84 , ii)0.56
Step-by-step explanation:
Given data
P(A) = professor arrives on time
P(A) = 0.8
P(B) = Student aarive on time
P(B) = 0.6
According to the question A & B are Independent
P(A∩B) = P(A) . P(B)
Therefore
& 
is also independent
= 1-0.8 = 0.2
= 1-0.6 = 0.4
part a)
Probability of both student and the professor are late
P(A'∩B') = P(A') . P(B') (only for independent cases)
= 0.2 x 0.4
= 0.08
Part b)
The probability that the student is late given that the professor is on time
= 
= 
= 0.4
Part c)
Assume the events are not independent
Given Data
P
= 0.4
=
= 0.4
= 0.4 x P
= 0.4 x 0.4 = 0.16
= 0.16
i)
The probability that at least one of them is on time
= 1-
= 1 - 0.16 = 0.84
ii)The probability that they are both on time
P
= 1 - 
= 1 - ![[P({A}')+P({B}') - P({A}'\cap {B}')]](https://tex.z-dn.net/?f=%5BP%28%7BA%7D%27%29%2BP%28%7BB%7D%27%29%20-%20P%28%7BA%7D%27%5Ccap%20%7BB%7D%27%29%5D)
= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56
