Lets call x the amount of 18% solution and y the amount of 40% solution, and write as equations the info of the problem:
18x + 40y = 10(20)
x + y = 10
lets multiply the second equation by -18 and add to the first:
18x + 40y = 200
-18x -<span> 18y = -180
</span>----------------------
0 + 22y = 20
y = 20/22 = 10/11
and substitute in the original equation:
x <span>+ y = 10
</span>x = 10 - y
x = 10 - 10/11
x = 110/11 - 10/11
x = 100/11
so they have to use 100/11 liters of 18% solution and 10/11 liters of 40% solution
The value of the 0 in the number 60,152 is the thousands place.
6 = ten thousand
0 = thousand
1 = hundreds
5 = tens
2 = ones
Hope helps!-Aparri
Answer:
<h2>There is 8% probability of Kitzen winning first and Ava second.</h2>
Step-by-step explanation:
The given table shows that there are 4 students, so there are 4 possible winner in total, but they have different number of tickets.
The total number of tickets is 48, that's the total possible outcomes. Kitzen has a probability of
Ava has a probability of
Now, the probability of having one event and the other is
Therefore, there is 8% probability of Kitzen winning first and Ava second.
(Notice that the probaility is not about Kitzen or Ava winning, it's about winning both, that's why the percentage is low)
Answer:
−35.713332 ; 0.313332
Step-by-step explanation:
Given that:
Sample size, n1 = 11
Sample mean, x1 = 79
Standard deviation, s1 = 18.25
Sample size, n2 = 18
Sample mean, x2 = 96.70
Standard deviation, s2 = 20.25
df = n1 + n2 - 2 ; 11 + 18 - 2 = 27
Tcritical = T0.01, 27 = 2.473
S = sqrt[(s1²/n1) + (s2²/n2)]
S = sqrt[(18.25^2 / 11) + (20.25^2 / 18)]
S = 7.284
(μ1 - μ2) = (x1 - x2) ± Tcritical * S
(μ1 - μ2) = (79 - 96.70) ± 2.473*7.284
(μ1 - μ2) = - 17.7 ± 18.013332
-17.7 - 18.013332 ; - 17.7 + 18.013332
−35.713332 ; 0.313332
