"1 indicating a coupon and all other outcomes indicating no coupon"
Probability is (number of successful outcomes) / (number of possible outcomes)
Theoretical Probability of rolling a 1: 1/8
Experimental Probability of using coupons: 4/48 = 1/12
So, the experimental probability of a customer using a coupon (that is, 1/12) is smaller than the theoretical probability of rolling a 1 (that is, 1/8).
Any line can be expressed in the form y=mx+b where m is the slope and b is y intercept.
Two lines can either be parallel ,overlap or meet at one point .Let us look at different cases :
1)When two lines are parallel they do not intersect at any point and hence the system of equations have no solution.
2) When two lines overlap each other then the two lines touch each other at infinite number of points and we say the system of equations have infinite solutions.
3) When two lines intersect each other at one point we say the system of equation has one solution.
Part A:
The given lines are intersecting at one point so we have one solution.
Part B:
The point of intersection is the solution to the system of equations .In the graph the point of intersection of the lines is (4,4)
Solution is (4,4)
The probability that the first two electric toothbrushes sold are defective is 0.016.
The probability of an event, say E occurring is:
Here,
n (E) = favorable outcomes
N = total number of outcomes
Let X = the number of defective electric toothbrushes sold.
The number of electric toothbrushes that were delivered to a store is n = 20.
The number of defective electric toothbrushes is x = 3.
The number of ways to select two toothbrushes to sell from the 20 toothbrushes is:
The number of ways to select two defective toothbrushes to sell from the 3 defective toothbrushes is:
Compute the probability that the first two electric toothbrushes sold are defective as follows:
P (Selling 2 defective toothbrushes) = Favorable outcomes ÷ Total no. of outcomes
Thus, the probability that the first two electric toothbrushes sold are defective is 0.016.
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Answer:
option D
Step-by-step explanation:
we have a right triangle
we know
c^2=a^2+b^2
c=hypotenuse
c=52 cm
a=48 cm
b= 20 cm
we substitute in the equation c^2=a^a+b^2
so we have
52^2=48^2+20^2
52= sqrt(48^2+20^2)
52= sqrt(2304+400)
52= sqrt(2704)
52cm=52cm
so c=52 cm
so the option D is the correct answer
