First, we’re going to use this formula:
2nd number - 1st number
————————————— x 100%
1st number
If we plug in the numbers to this equation:
23-20
———- x 100%
20
And subtract 23 to 20, the answer to this equation is:
3
—- x 100% = 15%
20
In this case, the percent of increase is 15%.
I hope this helps you
sec Q=1/cos Q
2=1/cos Q
cos Q= 1/2
Q=30+2.pi.n
Q=330+2.pi.n
n€Z
Given:
6 girls
4 boys
Probability of 1 girl being chosen: 1/6
Probability of 1 boy being chosen: 1/4
Probability of another boy being chosen: 1/3
1/6 = 0.17 x 100% = 17%
1/4 = 0.25 x 100% = 25%
1/3 = 0.33 x 100% = 33%
17% + 25% + 33% = 75%
The probability that exactly one girl and two boys will receive awards is 75%.
∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
----
∫(t = 2 to 3) t √(t - 2) dt
= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2
= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du
= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}
= 26/15.
----
For the k-entry, use integration by parts with
u = t, dv = sin(πt) dt
du = 1 dt, v = (-1/π) cos(πt).
So, ∫(t = 2 to 3) t sin(πt) dt
= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt
= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]
= 5/π + 0
= 5/π.
Therefore,
∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.
