Answer:
C. 1 / (a(a - 1))
Step-by-step explanation:
View Image
Just know that:
n! = n(n-1)!
= n(n-1)(n-2)!
= n(n-1)(n-2)(n-3)!
= ...
Answer: No, we don't have a right triangle
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Explanation:
If a triangle with sides a,b,c makes the equation a^2+b^2 = c^2 true, where c is the longest side, then this triangle is a right triangle. This is the converse of the pythagorean theorem.
Here we have a = 2, b = 5 and c = 7.
So...
a^2+b^2 = c^2
2^2+5^2 = 7^2
4+25 = 49
29 = 49
The last equation is false, so the first equation is false for those a,b,c values. Therefore, we do <u>not</u> have a right triangle.
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In contrast, consider the classic 3-4-5 right triangle
a = 3, b = 4 and c = 5 would make a^2+b^2 = c^2 true because 3^2+4^2 = 5^2 is a true equation (both sides lead to 25).
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
Answer:
Step-by-step explanation:
a) 3x^2-12x-11=3*x^2-3*2*2*x+3*4-23=3*(x^2-2*2x+2^2)-23=3*(x-2)^2-23
so a= -2 ,b= -23
b) x1=[12+V(12^2-4*3* -11)]/6=12/6+V(144+132)/6=2+2/6V69=2+1/3V69=
2+V69/9=2+V23/3
c=2, d= 23/3
