According to the probabilities given, it is found that the correct option regarding the independence of the events is given by:
No, P(carry cash) != P(carry cash|have children).
<h3>What is the probability of independent events?</h3>
If two events, A and B, are independent, we have that:

Which also means that:


In this problem, we have that:
- 62% carry cash on a regular basis, hence P(cash) = 0.62.
- 46% has children, hence P(children) = 0.46.
- Of the 46% who have children, 85% carry cash on a regular basis, hence P(cash|children) = 0.85.
Since P(carry cash) != P(carry cash|have children), they are not independent.
More can be learned about the probability of independent events at brainly.com/question/25715148
2(5-n)=-2+(-1*n)
2(5-n)=-n-2
distribute
10-2n=-n-2
add 2n to both sides
10=n-2
add 2 to both sides
12=n
Answer:
A) y - 2 = 3(x+4)
Step-by-step explanation:
Given that the equation is a line with slope =m=3 and
a point on it (x1,y1) = (-4,2)
Any line having slope m and passing through (x1,y1) has equation in point slope form as
y-y1 = m(x-x1)
We are given here x1 =-4, y1 =2 and m =3
y-y1 = y-2 and x-x1 = x-(-4) = x+4
So equation is y-2 = 3(x+4)
Verify:
The line is y = 3x+12+2 = 3x+14
Hence slope =3 is verified
Substitute the point x =-4 and y =2
2 =3(-4)+14 = 2 is true
Thus verified
Answer:
25
Step-by-step explanation:
Each of the 16 "ratio units" representing all 80 dancers must stand for ...
(80 dancers)/(16) = 5 dancers
Then 5 "ratio units" representing seventh-grade dancers will stand for ...
5 × 5 dancers = 25 dancers
___
Another way to figure this is to write the proportion ...
(seventh-grade dancers)/(all dancers) = 5/16
seventh-grade dancers = (all dancers)×(5/16) = 80×5/16
seventh-grade dancers = 25
Answer:

Step-by-step explanation:
Factorise numerator and denominator
x² - 9 ← is a difference of squares and factors in general as
a² - b² = (a - b)(a + b) , then
x² - 9
= x² - 3² = (x - 3)(x + 3)
x² - 3x ← factor out x from each term
= x(x - 3)
Then

=
← cancel common factor (x - 3) on numerator/denominator
= 