Use the sum-product pattern
2
−
−
1
2
x
2
−
x
−
12
x2−x−12
2
+
3
−
4
−
1
2
x
2
+
3
x
−
4
x
−
12
x2+3x−4x−12
2
Common factor from the two pairs
2
+
3
−
4
−
1
2
x
2
+
3
x
−
4
x
−
12
x2+3x−4x−12
(
+
3
)
−
4
(
+
3
)
x
(
x
+
3
)
−
4
(
x
+
3
)
x(x+3)−4(x+3)
3
Rewrite in factored form
(
+
3
)
−
4(+3)x
(x+3)−4(x+3)
x(x+3)−4(x+3)
(−4)(+3)
(x−4)(x+3)
(x−4)(x+3)
13
good job. really excelling :)
Kropot72
kropot72 3 years ago
This can be solved by using a standard normal distribution table. The z-score for 34 pounds is 1, the reason being that 34 is one standard deviation above the mean of 28 pounds.
Can use the table to find the cumulative probability for z = 1.00 and post the result? If you do this we can do the next simple steps.
Adult ticket (a) = $5
Child ticket (c) = $2
785 tickets = $3280
a + c = 785 tickets
5a + 2c = $3280
c = 215 child tickets
a = 570 adult tickets
570 + 215 = 785 tickets
5(570) + 2(215) = $3280
There were 215 child tickets sold on Saturday
Answer:
Try doing 4 eleven one half
Step-by-step explanation:
I not good at this sry
