Since you are given that the student registered early, the total number you deal with is all students who registered early.
211 + 329 = 540
number of undergraduates who registered early = 211
Among students who registered early:
p(undergraduate) = 211/540
Answer:
Step-by-step explanation:
-18.75: real, rational.
√36: real, rational, integer, whole, natural.
π²: real, irrational.
If x^2+bx+16 has at least one real root, then the equation x^2+bx+16=0 has at least one solution. The discriminant of a quadratic equation is b^2-4ac and it determines the nature of the roots. If the discriminant is zero, there is exactly one distinct real root. If the discriminant is positive, there are exactly two roots. The discriminant of <span>x^2+bx+16=0 is b^2-4(1)(16). The inequality here gives the values of b where the discriminant will be positive or zero:
b^2-4(1)(16) ≥ 0
</span><span>b^2-64 ≥ 0
(b+8)(b-8) </span><span>≥ 0
The answer is that all possible values of b are in the interval (-inf, -8]∪[8,inf) because those are the intervals where </span>(b+8)(b-8) is positive.
Answer:
The inequality is
10+7x\leq 4010+7x≤40
Step-by-step explanation:
Let
x-----> the number of child tickets
we know that
10+7x\leq 4010+7x≤40 -----> inequality that represent the situation
Solve for x
7x\leq 40-107x≤40−10
7x\leq 307x≤30
x\leq 4.3x≤4.3
The maximum number of child tickets is 4