To solve this problem, we make use of the Binomial
Probability equation which is mathematically expressed as:
P = [n! / r! (n – r)!] p^r * q^(n – r)
where,
n = the total number of gadgets = 4
r = number of samples = 1 and 2 (since not more than 2)
p = probability of success of getting a defective gadget
q = probability of failure = 1 – p
Calculating for p:
p = 5 / 15 = 0.33
So,
q = 1 – 0.33 = 0.67
Calculating for P when r = 1:
P (r = 1) = [4! / 1! 3!] 0.33^1 * 0.67^3
P (r = 1) = 0.3970
Calculating for P when r = 2:
P (r = 2) = [4! / 2! 2!] 0.33^2 * 0.67^2
P (r = 2) = 0.2933
Therefore the total probability of not getting more than
2 defective gadgets is:
P = 0.3970 + 0.2933
P = 0.6903
Hence there is a 0.6903 chance or 69.03% probability of
not getting more than 2 defective gadgets.
=Y2-10Y
We move all terms to the left:
-(Y2-10Y)=0
We add all the numbers together, and all the variables
-(+Y^2-10Y)=0
We get rid of parentheses
-Y^2+10Y=0
We add all the numbers together, and all the variables
-1Y^2+10Y=0
a = -1; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-1)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
Y1=−b−Δ√2aY2=−b+Δ√2a
Δ‾‾√=100‾‾‾‾√=10
Y1=−b−Δ√2a=−(10)−102∗−1=−20−2=+10
Y2=−b+Δ√2a=−(10)+102∗−1=0−2=0
Here the given expression is
We need to find the value of y when x =3 . So we substitute the value of x in the given expression.
So when x=3, y=0.4 and that's the required answer.
Using the distance formula
d= square root of (x2-x1)^2+(y2-y1)^2
(9-9)^2+(10-3)^2
0^2+7^2
0+49
Square root of 49 is 7
d=7
