★ Inequalities ★
or n ∈ ( 10 , 11 )
Possible value can be easily obtained by generating an arithmetic mean
Else we've infinite numbers between them ,
According to density property of real numbers , we can have any real number satisfying the given inequality under condition 10 < n < 11
Which is true for infinite possible numbers
10.1 , 10.2 , ... INFINITY
Answer:
0.3165
Step-by-step explanation:
This is a binomial distribution problem
Binomial distribution function is represented by
P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ
n = total number of sample spaces = number of adults sampled = 9
x = Number of successes required = number of selected adults that say they were too young to get tattoos = 0
p = probability of success = probability of an adult that regrets getting a tattoo saying they were too young to get the tattoo = 12% = 0.12
q = probability of failure = 1 - p = 1 - 0.12 = 0.88
P(X=0) = ⁹C₀ (0.12)⁰ (0.88)⁹⁻⁰
P(X=0) = 0.31647838183 = 0.3165
Hope this Helps!!!
<span>Exactly 8*pi - 16
Approximately 9.132741229
For this problem, we need to subtract the area of the square from the area of the circle. In order to get the area of the circle, we need to calculate its radius, which will be half of its diameter. And the diameter will be the length of the diagonal for the square. And since the area of the square is 16, that means that each side has a length of 4. And the Pythagorean theorem will allow us to easily calculate the diagonal. So:
sqrt(4^2 + 4^2) = sqrt(16 + 16) = sqrt(32) = 4*sqrt(2)
Therefore the radius of the circle is 2*sqrt(2).
And the area of the circle is pi*r^2 = pi*(2*sqrt(2)) = pi*8
So the area of the rest areas is exactly 8*pi - 16, or approximately 9.132741229</span>
Answer: 17
Steps:
1. Plug in (3) into “x” of the g(x) equation:
g(3) = (3)^2 + 4
g(3) = 9 +4
g(3) = 13
2. Plug in g(3) value into “x” of the f(x) equation:
f(g(3)) = x + 4
f(g(3)) = 13 + 4
f(g(3)) = 17
Answer:
6xy²z^4
Step-by-step explanation:
![\sqrt[3]{216x^3y^6z^12}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B216x%5E3y%5E6z%5E12%7D)
= ∛(6³)(x³)(y³)(y³)(z³)(z³)(z³)(z³) = 6x·y·y·z·z·z·z = 6xy²z^4
