The given equations are:
5x - 2y = 88
3x + 4y = 58
Multiplying the 1st equation by 2, we get the new set of equations as:
10x - 4y = 176
3x + 4y = 58
Adding the two equations, we get:
10x - 4y + 3x + 4y = 176 + 58
13x =234
x = 18
Using the value of x in 1st equation, we get:
5(18) - 2y = 88
- 2y = 88 -5(18)
-2y = -2
y = 1
So, the solution of the equation is (18, 1)
You can set up a system of equations for this problem. x= number of coach tickets and y = number of first class tickets.
$210x + $1200y = $10,230 (cost of coach ticket plus cost of first class tickets is total budget)
x + y = 11 (number of coach tickets plus number of first class tickets is total number of people)
Solve the second equation for y to get y = 11 - x, then plug that into the first equation and solve for x:
$210x + $1200(11 - x) = $10,230
$210x + $13,200 - $1200x = $10,230
-$990x + $13,200 = $10,230
-$990x = $2,970
x = 3
Sarah bought x = 3 coach tickets. Plug that into the second equation and solve for y:
3 + y = 11
y = 8
Sarah bought y = 8 first class tickets.
Answer: d. np(1 - p).
Step-by-step explanation:
Let x be any binomial variable which represents the number of success such that 
, where n is the sample size or the total number of trials and p is the probability of getting success in each trial .
Then, the mean E(x) and the variance Var(x) for the binomial distribution is given by equation :
where n is the sample size or the total number of trials and p is the probability of getting success in each trial .
Therefore , the correct option is option d. np(1 - p) .
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