1. a b c
2. a b c
3. a b c
4. a b c
5. a b c
then she eliminated 1 choice in 1 and 2, say as follows
1. b c
2. a b
3. a b c
4. a b c
5. a b c
Probability of answering correctly the first 2, and at least 2 or the remaining 3 is
P(answering 1,2 and exactly 2 of 3.4.or 5.)+P(answering 1,2 and also 3,4,5 )
P(answering 1,2 and exactly 2 of 3.4.or 5.)=
P(1,2,3,4 correct, 5 wrong)+P(1,2,3,5 correct, 4 wrong)+P(1,2,4,5 correct, 3 wrong)
also P(1,2,3,4 c, 5w)=P(1,2,3,5 c 4w)=P(1,2,4,5 c 3w )
so
P(answering 1,2 and exactly 2 of 3.4.or 5.)=3*P(1,2,3,4)=3*1/2*1/2*1/3*1/3*2/3=1/4*2/9=2/36=1/18
note: P(1 correct)=1/2
P(2 correct)=1/2
P(3 correct)=1/3
P(4 correct)=1/3
P(5 wrong) = 2/3
P(answering 1,2 and also 3,4,5 )=1/2*1/2*1/3*1/3*1/3=1/108
Ans: P= 1/18+1/108=(6+1)/108=7/108
Answer: 15x^2+6xy
Step-by-step explanation: 3x(6y−4y+5x)
(3x)(6y+−4y+5x)
(3x)(6y)+(3x)(−4y)+(3x)(5x)
18xy−12xy+15x^2
15x^2+6xy
Answer:
The first one
Step-by-step explanation:
For the first choice, the binomial is multiplied by itself, so it will result in a perfect square trinomial.
Answer:
1/3
Step-by-step explanation:
Answer:
Number of students in vans =15 and bus =46
Step-by-step explanation:
use variables for the vans and buses
let x= vans and y=buses
Now solve the 2 equations simultaneously
<em><u>remember</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>number</u></em><em><u> </u></em><em><u>your</u></em><em><u> </u></em><em><u>equations</u></em>
6x+7y=412..........1
6x+9y=504.........2
make x the subject of one equation
and label it equation 3
6x=412—7y

subs 3 into 2





subs 4 into 1





Therefore there are 15 students in each van and 46 students in each bus.