Answer:
P( sum is prime )= 73/216
Step-by-step explanation:
The minimum value of the sum will be 3 and maximum value will be 18. So the prime numbers in this range are 3 , 5, 7, 11, 13, 17.
P(sum=3)=1/216, P(sum=5)=6/216, P(sum=7)=15/216, P(sum=11)=27/216, P(sum=13)=21/216, P(sum=17)=3/216.
The final probability will be sum of the above given probabilities.
Hence P( sum is prime )= 73/216
Answer:
Step-by-step explanation:
The rectangular prism has a volume equal to V=xyz. V=(1/3)3(5/3)=5/3 in^3. The cube has a volume equal to V=s^3. The volume of the cube is equal to the prism when
![s^3=(1/3)(3)(5/3)\\ \\ s^3=5/3\\ \\ s=\sqrt[3]{\frac{5}{3}}in\\ \\ s\approx 1.19in](https://tex.z-dn.net/?f=s%5E3%3D%281%2F3%29%283%29%285%2F3%29%5C%5C%20%5C%5C%20s%5E3%3D5%2F3%5C%5C%20%5C%5C%20s%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B5%7D%7B3%7D%7Din%5C%5C%20%5C%5C%20s%5Capprox%201.19in)
79 is 4 below 83 so her next grade needs to be 4 above in order to get an average of 83. The answer is 87.
∫(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>.
The answer would be 54 since m=-2
Step one: 4(-2+5)+42
Step two: (-8+20)+42=54
