The best and most correct answer providedfrom your question about the entrance examination paper is the second option which is 2,250. The problem can be solved by:
3-1-1 : 5c3*5c1*5c1 *3!/2! = 750
<span>2-2-1 : 5c2*5c2*5c1 *3!/2! = 1500
</span>
Adding the two answers:
750 + 1500 = 2250
I hope it has come to your help.
Answer:
true
Step-by-step explanation:
Answer:
x=11
m<a=80º
m<b=80º
Step-by-step explanation:
They are congruent so
8x-8=5x+25
8x=5x+33
3x=33
x=11
Total tickets sold = 800
Total revenue = $3775
Ticket costs:
$3 per child,
$8 per adult,
$5 per senior citizen.
Of those who bought tickets, let
x = number of children
y = number of adults
z = senior citizens
Therefore
x + y + z = 800 (1)
3x + 8y + 5z = 3775 (2)
Twice as many children's tickets were sold as adults. Therefore
x = 2y (3)
Substitute (3) into (1) and (2).
2y + y + z = 800, or
3y + z = 800, or
z = 800 - 3y (4)
3(2y) + 8y + 5z = 3775, or
14y + 5z = 3775 (5)
Substtute (4) nto (5).
14y + 5(800 - 3y) = 3775
-y = -225
y = 225
From (4), obtain
z = 800 - 3y = 125
From (3), obtain
x = 2y = 450
Answer:
The number of tickets sold was:
450 children,
225 adults,
125 senior citizens.
Answer:
262/365
Step-by-step explanation:
So as you can see there is no more information aout Kay on her birthday, so the chances of her birthday being on a week day is given by the total number of the weekdays of the year between the total number of days in a year, so in 2019 there are 262 weekdays, divided by 365 you get the probability that Kay´s birthday falls on a weekday.
262/365=,7178=71,78%
So the probability of Kay´s brithday falling on a week day will be 71,72%
