I’m sorry this is messy but I hope this is right!! <3
<em>Greetings from Brasil...</em>
The average for a set of 9 elements will be
(A + B + C + D + E + F + G + H + I) ÷ 9 = 20
Let's make (A + B + C + D + E + F + G + H + I) like S
<em>(I chose S to remember a sum)</em>
Let us think.....
S ÷ 9 = 20
S = 20 × 9
S = 180
So, (A + B + C + D + E + F + G + H + I) = 180
According to the statement, we will include a number (element J) in the sum to obtain a mean of (20 - 4), that is:
<h3>(A + B + C + D + E + F + G + H + I +
J) ÷ 10 = (20 - 4)</h3>
as seen above, (A + B + C + D + E + F + G + H + I) = 180, then
(180 + J) ÷ 10 = 16
(180 + J) = 160
J = 160 - 180
<h2>J = - 20</h2><h2 />
So, including the number - 20 <em>(minus 20)</em> in the original mean we will obtain a new mean whose result will be 16
Y = 4 and x = 1 hope this helps!
Answer:
The probability that a student is proficient in mathematics, but not in reading is, 0.10.
The probability that a student is proficient in reading, but not in mathematics is, 0.17
Step-by-step explanation:
Let's define the events:
L: The student is proficient in reading
M: The student is proficient in math
The probabilities are given by:
The probability that a student is proficient in mathematics, but not in reading is, 0.10.
The probability that a student is proficient in reading, but not in mathematics is, 0.17
Answer:
16. r(s)(t) =
r=2, s=3, t=4
(2)(3)(4)=
(2*3) = 6 ---> 6(4) = 24
final: 24
17. (r)(s)(t)(x)(y) =
r = 2, s = 3, t = 4, x = 5, y = 6
(2)(3)(4)(5)(6)
2*3=6
6*4=24
24*5= 120
120*6=720
final: 720
18. (7x + 2Y) =
x = 5, y = 6
rewrite---> = [7(5) + 2(6)]
[35 + 12] = 47
final: 47
hope this helpsss sorry took a while LOL
Step-by-step explanation:
q = 1; r = 2; s = 3; t = 4; x = 5; y = 6
