The cost of the trip was $56.25
Answer:
When we have a rational function like:

The domain will be the set of all real numbers, such that the denominator is different than zero.
So the first step is to find the values of x such that the denominator (x^2 + 3) is equal to zero.
Then we need to solve:
x^2 + 3 = 0
x^2 = -3
x = √(-3)
This is the square root of a negative number, then this is a complex number.
This means that there is no real number such that x^2 + 3 is equal to zero, then if x can only be a real number, we will never have the denominator equal to zero, so the domain will be the set of all real numbers.
D: x ∈ R.
b) we want to find two different numbers x such that:
r(x) = 1/4
Then we need to solve:

We can multiply both sides by (x^2 + 3)


Now we can multiply both sides by 4:


Now we only need to solve the quadratic equation:
x^2 + 3 - 4*x - 4 = 0
x^2 - 4*x - 1 = 0
We can use the Bhaskara's formula to solve this, remember that for an equation like:
a*x^2 + b*x + c = 0
the solutions are:

here we have:
a = 1
b = -4
c = -1
Then in this case the solutions are:

x = (4 + 4.47)/2 = 4.235
x = (4 - 4.47)/2 = -0.235
Answer:
10
Step-by-step explanation:
Answer:
Let X be the event of feeling secure after saving $50,000,
Given,
The probability of feeling secure after saving $50,000, p = 40 % = 0.4,
So, the probability of not feeling secure after saving $50,000, q = 1 - p = 0.6,
Since, the binomial distribution formula,

Where, 
If 8 households choose randomly,
That is, n = 8
(a) the probability of the number that say they would feel secure is exactly 5



(b) the probability of the number that say they would feel secure is more than five




(c) the probability of the number that say they would feel secure is at most five




Answer:
Red: 1/5 or 0.2 or 20% (depends if the question asked you to give the answer in fraction/decimal/percentage. Either way, it's still correct)
Black: 0
Blue: 2/5 or 0.4 or 40%
Step-by-step explanation:
In total, the bag contains 10 balls,
- There are no black marbles so you have 0 chance of pulling them out.