7) (12 + 6)/(2 +4)
(18)/(6) = 3
8) (42 - 24)/6
(18)/6 = 3
9) (9 + 16) - (2 x 4)
25 - 8 = 17
10) 60 - (3 + 2) x 5
60 - 5 x 5
60 - 25 = 35
11) 18 + (9/3)
18 + 3 = 21
12) 5 x (2 + 4 + 3)
5 x (9) = 45
hope this helps
∫(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>.
Answer:
1/8 :)
Step-by-step explanation:
1/8 is correct because 12.5*8=100
The GCF if 3. Please mark Brainliest!!!
Answer:
the numerical value of the correlation between percent of classes attended and grade index is r = 0.4
Step-by-step explanation:
Given the data in the question;
we know that;
the coefficient of determination is r²
while the correlation coefficient is defined as r = √(r²)
The coefficient of determination tells us the percentage of the variation in y by the corresponding variation in x.
Now, given that class attendance explained 16% of the variation in grade index among the students.
so
coefficient of determination is r² = 16%
The correlation coefficient between percent of classes attended and grade index will be;
r = √(r²)
r = √( 16% )
r = √( 0.16 )
r = 0.4
Therefore, the numerical value of the correlation between percent of classes attended and grade index is r = 0.4
