Guys please help with this question thanks

The probability that a customer's order is not shipped on time is 0.06. A particular customer places three orders, and the order

charle [14.2K]

Answer:

a) There is a 83.06% probability that all orders are shipped on time.

b) There is a 15.90% probability that exactly one order is not shipped ontime.

c) The probability of at least two orders being late is 1.02% + 0.02% = 1.04%.

Step-by-step explanation:

Probability:

What you want to happen is the desired outcome.

Everything that can happen iis the total outcomes.

The probability is the division of the number of possible outcomes by the number of total outcomes.

In our problem, there is:

-A 6% probability that a customer's order is not shipped on time.

-A 94% probability that a customer's order is shipped on time.

We have these following orders:

O1 - O2 - O3.

(a) What is the probability that all are shipped on time?

The probabilities that each order is shipped on time are O1 = 0.94, O2 = 0.94 and O3 = 0.94. So:

$P = (0.94)^{3}$ = 0.8306

There is a 83.06% probability that all orders are shipped on time.

(b) What is the probability that exactly one is not shipped ontime?

The order's can be permutated. What this means? It means that we can have O1 late and O2,03 on time, O2 late and O1,O3 on time and O3 late and O1, O2 on time. We have a permutation of 3 elements(the orders) with 2 and 1 repetitions(2 on time and one late).

The probability that an order is late is:

$P = (0.94)^{2}(0.06)$ = 0.053 for each permutation

Considering the permutations:

$P = 0.053*p^{3}_{2,1} = 0.053\frac{3!}{2!*1!} = 0.053*3 = 0.1590$

There is a 15.90% probability that exactly one order is not shipped ontime.

(c) What is the probability that two or more orders are not shipped on time?

P = P1 + P2, where P1 is the probability that two orders are late and P3 is the probability that all three orders are late.

P1

Considering the permutations, the probability that two orders are late is:

$P_{1} = p^{3}_{2,1}*(0.94)*(0.06)^{2} = 3*(0.94)*(0.06)^{2} = 0.0102$

There is a 1.02% probability that two orders are late

P2

$P_{2} = (0.06)^3 = 0.0002$

There is a 0.02% probability that all three orders are late.

The probability of at least two orders being late is 1.02% + 0.02% = 1.04%.

5 0

3 years ago

Answer please////////

BartSMP [9]

Answer:

Coordinate of B: (3, -5)

Coordinate of C: (3, 5)

Coordinate of D: (0, -1)

Step-by-step explanation:

Point A is (-2, -5).

The distance between line AB is 5 units.

The distance between line BC is 10 units.

To find the point for B, the distance is 5 units on the x axis.

The x point for point B is -2 + 5 which is 3 units. x = 3

The y point doesn't change, so the coordinate of point B is (3, -5).

___________________________________________________________

The distance between line BC is 10 units.

To find the point of C, the distance is 10 unites on the y axis.

The y point for point b is -5 + 10 which is 5 units. y = 5

The x point doesn't change, so the coordinate of point C is (3, 5).

___________________________________________________________

We find an equation for line AC. The equation is y = mx + b.

Y is any y point on the line. X is any x point on the line. M is the slope of the line. B is the y-intercept of the line.

Since D is right on the y axis, this means that the x point of D is 0.

Point C is (3, 5) and point A is (-2, -5).

We use (y2 - y1) / (x2 - x1) and substitute. Y is the y point and x is the x point.

(5 - (-5)) / (3 - (-2)) = 10 / 5 = 2

The slope of the line is 2.

With this, we get the equation of y = 2x + b

We insert one of the random x point and y point which (in this case) I choose (3, 5).

Substituting, we get -

5 = 2 * 3 + b

5 = 6 + b

b = 5 - 6

b = -1

So the y-intercept of the graph is -1 and for the graph, we get -

y = 2x - 1

D is already the line of the y-intercept, but we can make sure by inserting the x point as 0.

y = 2 * 0 - 1

y = 0 - 1

y = -1

The coordinate of point D is (0, -1).

8 0

3 years ago

A real estate agent has 17 properties that she shows. She feels that there is a 60% chance of selling any one property during a

Zarrin [17]

Answer:

$P(X\leq 5)$

And we can find the individual probabilities like this:

$P(X=0)=(17C0)(0.6)^0 (1-0.6)^{17-0}=0.000000171$

$P(X=1)=(17C1)(0.6)^1 (1-0.6)^{17-1}=0.00000439$

$P(X=2)=(17C2)(0.6)^2 (1-0.6)^{17-2}=0.0000526$

$P(X=3)=(17C3)(0.6)^3 (1-0.6)^{17-3}=0.000394$

$P(X=4)=(17C4)(0.6)^4 (1-0.6)^{17-4}=0.00207$

$P(X=5)=(17C5)(0.6)^5 (1-0.6)^{17-5}=0.000807$

And adding we got:

$P(X\leq 5) = 0.0106$

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: $P(A)+P(A') =1$

Solution to the problem

Let X the random variable of interest, on this case we know that:

$X \sim Binom(n=17, p=0.6)$

And we want this probability:

$P(X\leq 5)$

And we can find the individual probabilities like this:

$P(X=0)=(17C0)(0.6)^0 (1-0.6)^{17-0}=0.000000171$

$P(X=1)=(17C1)(0.6)^1 (1-0.6)^{17-1}=0.00000439$

$P(X=2)=(17C2)(0.6)^2 (1-0.6)^{17-2}=0.0000526$

$P(X=3)=(17C3)(0.6)^3 (1-0.6)^{17-3}=0.000394$

$P(X=4)=(17C4)(0.6)^4 (1-0.6)^{17-4}=0.00207$

$P(X=5)=(17C5)(0.6)^5 (1-0.6)^{17-5}=0.000807$

And adding we got:

$P(X\leq 5) = 0.0106$

5 0

3 years ago

Find the zeros of the function f(x) = 1/3 x^2 + 72

Alexus [3.1K]

To find the zeros of an equation equate the equation to zero

Please with this question I don’t really get it looking at how u hv written it
Is the x^2 a numerator or a denominator
I don’t really get how u wrote ur question

6 0

3 years ago

jared needs to drill holes 2 1/4 inches apart in a 36 inch section of his garage. which expression can he use to figure out the

MrRissso [65]

Answer: 36 / 2 1/4

Step-by-step explanation:

From the question, we are informed that Jared needs to drill holes 2 1/4 inches apart in a 36 inch section of his garage. The expression that he can use to figure out the number of holes he will need to drill will be:

= 36 ÷ 2 1/4

From the expression we can then calculate the number of hours to drill. This will be:

= 36 / 2 1/4

= 36 / 9/4

= 36 × 4/9

= 16

He'll need to drill 16 holes

8 0

3 years ago

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