Answer:
No. of boxes = 10
Oranges per box = 56
Total no. of oranges = 56*10=560
No. of bad oranges = 560/40= 14
(i) Probability of bad orange = 14/560 = 1/40
(ii) no of oranges expected to be bad = 14 ( found above)
Hope it helps.............. :)
Answer:
1. reflection across x-axis
2. translation 6 units to the right and 3 units up (x+6,y+3)
Step-by-step explanation:
The trapezoid ABCD has it vertices at points A(-5,2), B(-3,4), C(-2,4) and D(-1,2).
First transformation is the reflection across the x-axis with the rule
(x,y)→(x,-y)
so,
- A(-5,2)→A'(-5,-2)
- B(-3,4)→B'(-3,-4)
- C(-2,4)→C'(-2,-4)
- D(-1,2)→D'(-1,-2)
Second transformation is translation 6 units to the right and 3 units up with the rule
(x,y)→(x+6,y+3)
so,
- A'(-5,-2)→E(1,1)
- B'(-3,-4)→H(3,-1)
- C'(-2,-4)→G(4,-1)
- D'(-1,-2)→F(5,1)
H is the correct answer. Y= (x+2)^2-5
2 to the left x+2=0 x=-2
5 dpwn
Answer:
Let's define the variables:
A = price of one adult ticket.
S = price of one student ticket.
We know that:
"On the first day of ticket sales the school sold 1 adult ticket and 6 student tickets for a total of $69."
1*A + 6*S = $69
"The school took in $150 on the second day by selling 7 adult tickets and student tickets"
7*A + 7*S = $150
Then we have a system of equations:
A + 6*S = $69
7*A + 7*S = $150.
To solve this, we should start by isolating one variable in one of the equations, let's isolate A in the first equation:
A = $69 - 6*S
Now let's replace this in the other equation:
7*($69 - 6*S) + 7*S = $150
Now we can solve this for S.
$483 - 42*S + 7*S = $150
$483 - 35*S = $150
$483 - $150 = 35*S
$333 = 35*S
$333/35 = S
$9.51 = S
That we could round to $9.50
That is the price of one student ticket.
35/8 = 4.376 5/3 = 1.666… 4.376/1.666…
Therefore 2.625 pieces