7/10
3 are striped
total = 10
not = 70%
There are 10 marbles 7 of them are not striped/swirled so that makes it 6/10 marbles are not striped/swirled so the answer would be either 7/10, 70%, 0.7 (there are the same thing)
10 marbles 3 are striped 7 are not 7 out of 10 7/10
The expression simplified to 32.
Your answer is wrong :(
You would expect 3 because you have an 80% chance of drawing a 3 but you have a 20% chance of drawing a 9 so you would get 3
1. Three segments have a total length of 26 inches. The first segment is 2 less than three times the second segment. The third segment is 1 more than five times the second segment. What are the lengths of the three segments?
All three variables must be defined in terms of x.
1st 3x - 2
2nd x
3rd 5x + 1
3x - 2 + x + 5x + 1 = 26
9x = 27
x = 3
18 / 30 = x / 24 or x = (24 * 18) / 30
30x = 24 * 18 x = 432 / 30
x = 14.4 x = 14.4
9x4 is the same as 9+9+9+9=36 or 4+4+4+4+4+4+4+4+4=36
The numbers, but in word form.
The answer is 13211Q
1. Q
2. 1Q One Q (the previous , but in word form)
3. 111Q, One 1, One Q (11,1Q)
4. 311Q Three one(s), One Q (31,1Q)
Ans: 13211Q One three, 2 one, one Q (13, 21, 1Q)
thank you! I knew how to do this part, I got b too, thank you :)
The answer is 10.
Those towers honestly looks like my hand pointing up.
1. Almost right; you forgot negative values
(just as a reminder, the sum \(\displaystyle\sum_{n=1}^{\infty}x^n\) where x is a constant only converges when \(-1 by the ratio test.)
So the answer would be \((-\infty, -2)\bigcup(2, \infty)\)
2. The sum \(\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{y}{n}\right)^a\), where y is a constant, only converges when \(a>1\), so the answer is \((1, \infty)\)
A. H (0, 4, 2) D (0, 4, 0) G (5, 4, 2) F (5, 0, 2) I don't know about B
Thank you!!!
So in the question, it says it can't start or end with a vowel and the only vowel there is I and out of the 5 letters _ _ _ _ _ we now know that there are 4 possibilities for the 1st and last letter and now there can be 5 each in the middle so our equation is 4*5*5*5*4 which is equal to 2000
9-(1/2)4×48=87
sorry, none of those are correct :(
Note that we can write this as
C = 2n^(1/4) [ 20n^(1/8) + 8n^(1/3) ]
C = 2 * 20* n^(1/4 + 1/8) + 2* 8 * n^(1/4 + 1/3)
C = 40 n^(3/8) + 16 n^(7/12)
C = 40 8√ (n3) + 16 12√ (n7)
a = 3
b = 7
Let A be your current age
(4/5)A > (3/4) (A + 1)
(4/5)A - (3/4)A > 3/4
(1/20)A > 3/4
A > (3/4)(20)
A > 15
And
(4/5)A > (5/6) (A - 2)
(4/5)A > (5/6)A - 5/3
5/3 > (5/6)A - (4/5)A
5/3 > (1/30)A
(5/3) ( 30) > A
50 > A
15 < A < 50 where A is an integer
Wait... nvm. There's no answer to it.
https://web2.0calc.com/questions/four-fifths-of-my-current-age-is-greater-than-three-quarters-of-my-age-one-year-from-now
Um... CPhill answered this question a couple of years ago, if you want to see the answer.
Area = (1/2) height * ( sum of base lengths)
38 = (1/2) (4) ( 10 + x)
38 = 2 ( 10 + x)
38 / 2 = 10 + x
19 = 10 + x
19 - 10 = x = 9 = the other base length
In every hundred, we will have 20 integers that satisfy this
So 20 (hundreds) * 20 ( occurnces in each hundred) = 400
And we have three more occurences from 2001 - 2012
So
400 + 3 =
403 occurences