Well, I don't know. Let's see . . . . .
First draw: One 'M' available out of 11 letters. Probability of picking it = 1/11.
2nd draw: Four 'I's available out of 10 letters. Probability of picking one = 4/10.
3rd draw: Four 'S's available out of 9 letters. Probability of picking one = 4/9.
4th draw: Three 'S's available out of 8 letters. Probability of picking one = 3/8.
5th draw: Three 'I's available out of 7 letters. Probability of picking one = 3/7.
6th draw: Two 'S's available out of 6 letters. Probability of picking one = 2/6.
7th draw: One 'S' available out of 5 letters. Probability of picking it = 1/5.
8th draw: Two 'I's available out of 4 letters. Probability of picking it = 2/4.
9th draw: Two 'P's available out of 3 letters. Probability of picking one = 2/3.
10th draw: One 'P' available out of 2 letters. Probability of picking it = 1/2.
11th draw: One letter left. It is an 'I'. Probability of picking it = 1 .
Probability of all of those draws in order =
(1/11) x (4/10) x (4/9) x (3/8) x (3/7) x (2/6) x (1/5) x (2/4) x (2/3) x (1/2) x (1) =
1,152 / 39,916,800 =
1 / 34,650 =
0.00002886 =
<em> 0.002886 percent</em> (rounded)
Not a good bet. (But better than the lottery.)
Multiply the length and height of the paper then divide that by 3 or 3^2 i get confused on that part but pick the answer that makes the most sence
9514 1404 393
Answer:
3. $6000 at 8%; $5000 at 5%
4. 66 2/3 gallons of 9%; 133 1/3 gallons of 6%
Step-by-step explanation:
For many mixture problems, it works well to let a variable represent the amount of the largest contributor used in the mix.
3. Let x represent the amount invested at 8%. Then the total interest is ...
8%x +5%(11000-x) = 730
3%x = 730 -550
x = 180/0.03 = 6000
$6000 was invested at 8%; $5000 was invested at 5%.
__
4. Let x represent the number of gallons of 9% content. Then the alcohol content of the mix is ...
9%x +6%(200 -x) = 7%(200)
3%x = 1%(200) . . . . . subtract 6%(200)
x = 200/3 = 66 2/3 . . . . . divide by 3%
200 -x = 133 1/3
66 2/3 gallons of 9% wine and 133 1/3 gallons of 6% wine should be used.
My best guess is <span>(r): 0.91</span>
