I do believe that it is C because the points fall above the value of y, and I do believe that the sign given here is supposed to be shown by a solid black line, meaning that points on the graph can be shown on the line.
Answer:
<h2><u>question 1</u></h2>
use product and sum method
product = -96
sum = -20
numbers needed = ( -24 , 4)
n - 24 = 0
n + 4 = 0
hence <u>n = 24 and n = -4 </u>
<u></u>
<h2><u>Question 2 </u></h2>
<u />
<u />
in the form 
= 
make use of the formula :
replace values to make 2 equations :
1.
= 3.17
2.
= -15.2
hence <u>x = 3.17 and x = -15.2</u>
<u />
<h2><u>Question 3 </u></h2>
<u />
<u />
use product and sum method
product = 40
sum = -14
numbers needed = (-10 , -4)
x - 10 = 0
x - 4 = 0
hence<u> x = 10 and x = 4</u>
<u />
<h2><u>Question 4 </u></h2>
<u />
<u />
in the form
this becomes 
= 
can simplify by 5
= 
use product and sum method
product = -5
sum = -4
numbers needed (-5 , 1)
b-5 = 0
b + 1 = 0
hence <u>b = 5 and b = -1</u>
Answer:
a) 1/2
b) 1/n
c) 1/4
Step-by-step explanation:
a) For each permutation, either 1 precedes 2 or 2 precedes 1. For each permutation in which 1 precedes 2, we can swap 1 and 2 to obtain a permutation in which 2 preceds 1. Thus, half of the total permutations will involve in 1 preceding 2, hence, the probability for a permutation having 1 before 2 is 1/2.
c) If 2 is at the start of the permutation, then it is impossible for 1 to be before 2. If that is not the case, then 1 has a probability of 1/n-1 to be exactly in the position before 2. We can divide in 2 cases using the theorem of total probability,
P( 1 immediately preceds 2) = P (1 immediately precedes 2 | 2 is at position 1) * P(2 is at position 1) + P(1 immediately precedes 2 | 2 is not at position 1) * P(2 is not at position 1) = 0 * 1/n + (1/n-1)*(n-1/n) = 1/n.
d) We can divide the total of permutations in 4 different groups with equal cardinality:
- Those in which n precedes 1 and n-1 precedes 2
- those in which n precedes 1 and 2 precedes n-1
- those in which 1 precedes n and n-1 precedes 2
- those in which 1 precedes n and 2 precedes n-1
All this groups have equal cardinality because we can obtain any element from one group from another by making a permutations between 1 and n and/or 2 and n-1.
This means that the total amount of favourable cases (elements of the first group) are a quarter of the total, hence, the probability of the event is 1/4.
Answer:
1,123.30
Step-by-step explanation:
first of all, you are decreasing 1015 by 6 percent so it will be
1015 * 0.93 because you are decreasing it from the percentage
then, you are increasing it by 19 percent so you mulitply what you got on the previous calculation and multiply it by 1.19
943.95 * 1.19
and answer will be 1,123.30
Answer:
rad 3
Step-by-step explanation:
it just is
