Answer:
x = 7
Step-by-step explanation:
y = 14
14 x 2 = 28
Therefore:
3.5 x 2 = 7
The number of presale tickets sold is 271
<em><u>Solution:</u></em>
Let "p" be the number of presale tickets sold
Let "g" be the number of tickets sold at gate
<em><u>Given that, total of 800 Pre-sale tickets and tickets at the gate were sold</u></em>
Therefore,
Presale tickets + tickets sold at gate = 800
p + g = 800 ------ eqn 1
<em><u>Given that, number of tickets sold at the gate was thirteen less than twice the number of pre-sale tickets</u></em>
Therefore,
Number of tickets sold at gate = twice the number of pre-sale tickets - 13
g = 2p - 13 ------- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
Substitute eqn 2 in eqn 1
p + 2p - 13 = 800
3p -13 = 800
3p = 800 + 13
3p = 813
p = 271
Thus 271 presale tickets were sold
Answer:
Pr(X >42) = Pr( Z > -2.344)
= Pr( Z< 2.344) = 0.9905
Step-by-step explanation:
The scenario presented can be modeled by a binomial model;
The probability of success is, p = 0.65
There are n = 80 independent trials
Let X denote the number of drivers that wear a seat belt, then we are to find the probability that X is greater than 42;
Pr(X > 42)
In this case we can use the normal approximation to the binomial model;
mu = n*p = 80(0.65) = 52
sigma^2 = n*p*(1-p) = 18.2
Pr(X >42) = Pr( Z > -2.344)
= Pr( Z< 2.344) = 0.9905
