9. 12 edges
Work: <u>4</u> horizontal edges on top, <u>4</u> horizontal edges on bottom, <u>4</u> vertical edges for sides
10. Ten-thousands place
Work: The 8 is highlighted. The order of places is: <u>ten-thousands</u>, thousands, hundreds, tens, ones
11. 10 4/5
Work: 54 ÷ 5 ——> 50÷5 = <u>10</u> with 4 left over and put into a fraction of <u>4/5</u>
12. 3,000
Work: 2 is in the thousands place so you look to the place behind it and there is an 8. If the number behind is 5 through 9 then you round up. 8 is above 5, therefore, you round the 2 up to a <u>3</u>.
13. 0.45
Work: The pattern is +0.03 so 0.42+0.03=<u>0.45</u>
14. 54 cakes
Work: <u>6 eggs</u> per 1 cake • <u>9 cakes</u> = 6•9=54 eggs used
15. 378 desks in the school
Work: 6 rows • 7 desks per row = 42 desks in 1 classroom. 42 desks • 9 classrooms = 378 desks in the whole school
Yes it is a solution because if you take the equation and plug the point in for the x and y :
4 (2) + 3 (3) < 20
8 + 9 < 20
17 < 20
which is true
10 (2) + (3) > 10
20 + 3> 10
23>10
which is also true
Answer:
(a) 0.28347
(b) 0.36909
(c) 0.0039
(d) 0.9806
Step-by-step explanation:
Given information:
n=12
p = 20% = 0.2
q = 1-p = 1-0.2 = 0.8
Binomial formula:
(a) Exactly two will be drunken drivers.
Therefore, the probability that exactly two will be drunken drivers is 0.28347.
(b)Three or four will be drunken drivers.
Using binomial we get
Therefore, the probability that three or four will be drunken drivers is 0.3691.
(c)
At least 7 will be drunken drivers.
![P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]](https://tex.z-dn.net/?f=P%28x%5Cleq%207%29%3D1-%5BP%28x%3D0%29%2BP%28x%3D1%29%2BP%28x%3D2%29%2BP%28x%3D3%29%2BP%28x%3D4%29%2BP%28x%3D5%29%2BP%28x%3D6%29%5D)
![P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]](https://tex.z-dn.net/?f=P%28x%5Cleq%207%29%3D1-%5B0.06872%2B0.20616%2B0.28347%2B0.23622%2B0.13288%2B0.05315%2B0.0155%5D)
![P(x\leq 7)=1-[0.9961]](https://tex.z-dn.net/?f=P%28x%5Cleq%207%29%3D1-%5B0.9961%5D)
Therefore, the probability of at least 7 will be drunken drivers is 0.0039.
(d) At most 5 will be drunken drivers.
Therefore, the probability of at most 5 will be drunken drivers is 0.9806.
So,
You have to undistribute a common factor in all of the terms.
2a + 2b
There is a 2 in each term.
Undistribute.
2(a + b)
