Antawn Jamison is shooting free throws. Making or missing free throws doesn't change the probability that

Every day, a lecture may be canceled due to inclement weather with probability 0.05. Class cancelations on different days are in

andrezito [222]

<h3>(a)Probability of cancelling at least 4 classes is 0.0055.</h3><h3>(b)Probability of 10th class is third class that gets cancelled = 0.0031</h3>

Step-by-step explanation:

Here, the question is<u> incomplete</u>.

Every day, a lecture may be canceled due to inclement weather with probability 0.05. Class cancellations on different days are independent.

(a) There are 15 classes left this semester. Compute the probability that at least 4 of them get canceled.

(b) Compute the probability that the tenth class this semester is the third class that gets canceled.

Now, here:

The probability of cancelling each class = 0.05

Now, probability of cancelling at least 4 classes

= 1 - P(Cancelling at max 3 classes)

= 1 - P(0 ≤ x ≤ 3) = 1 - Binomial (15,0.05,3)

= 0.0055

Hence, probability of cancelling at least 4 classes is 0.0055.

(b) As given, 10th class is third class that gets cancelled.

So, the first and second classes that get cancelled in between 1 - 9.

P(2 class cancelled in 1st 9) = Binomial (15,0.05,3) = 0.0629

Also, as given P(10th class cancelled) = 0.05

⇒ P(10th class is third class that gets cancelled ) = 0.0629 x 0.05 = 0.0031

5 0

1 year ago

What value of x would make KM ∥ JN?

stepan [7]

Answer:

x = 10

Step-by-step explanation:

By the converse of the side-splitter theorem, if $\dfrac{JK}{KL}= \dfrac{NM}{ML}$ , then KM ∥ JN.

Substitute the expressions into the proportion:

$\dfrac{x-5}{x}= \dfrac{x-3}{x+4}$

Cross-multiply:

(x – 5)(x+4) = x(x – 3).

Distribute:

$x(x) + x(4) - 5(x) - 5(4) = x(x) + x(-3).$

Multiply and simplify:

$x^2 +4x- 5x - 20 = x^2-3x\\-x-20=-3x\\$Collect like terms$\\-x+3x=20\\2x=20\\$Divide both sides by 2\\x=10$

Solve for x: x = 10

Therefore, the value of x that would make KM parallel to JN is 10.

4 0

1 year ago

The number of defective components produced by a certain process in one day has a Poisson distribution with a mean of 20. Each d

Kruka [31]

Answer:

The probability that exactly 15 defective components are produced in a particular day is 0.0516

Step-by-step explanation:

Probability function : $P(X=x)=e^{-\lambda} \frac{\lambda^x}{x!}$

We are given that The number of defective components produced by a certain process in one day has a Poisson distribution with a mean of 20.

So, $\lambda = 20$

we are supposed to find the probability that exactly 15 defective components are produced in a particular day

So,x = 15

Substitute the values in the formula :

$P(X=15)=e^{-20} \frac{20^{15}}{15!}$

$P(X=15)=0.0516$

Hence the probability that exactly 15 defective components are produced in a particular day is 0.0516

8 0

1 year ago

Tiny's Tattoo Parlor sold 389 tattoos this month. 43 of those tattoos were arm tattoos. Based on this data, what is a reasonable

antiseptic1488 [7]

Answer
(389-43)/389=346/389

8 0

1 year ago

Read 2 more answers

When you select a random person, the probability that this person will go to Ches

insens350 [35]

Answer: a= 0.0231

Step-by-step explanation:

n=8

p=0.75

q=1 - 0.75 = 0.25

(p=x)

p(5) = 8C5 (0.25)^5 (0.75)^8-5

p(5)= 0.0231

Or 2.31%

6 0

1 year ago