If I'm reading this correctly, it is 2*3*n*n = 6n^2
6n^2 = 54 Divide by 6
n^2 = 54/6
n^2 = 9 Take the square root of both sides.
sqrt(n^2) = sqrt(9)
n = +/- 3
n = + 3 or n = -3. Both will solve the equation above.
Edit
2 * 3 * n * n
Let n = 3
2 * 3 * 3 * 3 = 6 * 9 = 54
The other value is n = - 2
2 * 3 * n * n
2 * 3 * (-3) * (-3) A minus times a minus is a plus.
2 * 3 * 9 = 6 * 9 = 54
both + 3 and - 3 for n will give 54.
4b^2 - 16 =
4(b^2 - 4) =
4(b + 2)(b - 2)
Answer:
Points C,D, E and F are coplanar.
Answer:
x=3
Step-by-step explanation:
please mark me as brainliest
Answer:
P(≥ 7 males) = 0.0548
Step-by-step explanation:
This is a binomial probability distribution problem.
We are told that Before 1918;
P(male) = 40% = 0.4
P(female) = 60% = 0.6
n = 10
Thus;probability that 7 or more were male is;
P(≥ 7 males) = P(7) + P(8) + P(9) + P(10)
Now, binomial probability formula is;
P(x) = [n!/((n - x)! × x!)] × p^(x) × q^(n - x)
Now, p = 0.4 and q = 0.6.
Also, n = 10
Thus;
P(7) = [10!/((10 - 7)! × 7!)] × 0.4^(7) × 0.6^(10 - 7)
P(7) = 0.0425
P(8) = [10!/((10 - 8)! × 8!)] × 0.4^(8) × 0.6^(10 - 8)
P(8) = 0.0106
P(9) = [10!/((10 - 9)! × 9!)] × 0.4^(9) × 0.6^(10 - 9)
P(9) = 0.0016
P(10) = [10!/((10 - 10)! × 10!)] × 0.4^(10) × 0.6^(10 - 10)
P(10) = 0.0001
Thus;
P(≥ 7 males) = 0.0425 + 0.0106 + 0.0016 + 0.0001 = 0.0548
