All categories

3 years ago

Plz help, no games its already overdue and has to get done!

Mathematics

2 answers:

Finger [1]3 years ago

8 0

1: True

2: True

3: False

4: True

ZanzabumX [31]3 years ago

7 0

Answer:

True True False True

Step-by-step explanation:

You might be interested in

Someone please help me with this question

alexira [117]

Is there another part of this?

5 0

3 years ago

Find the unit rate. $121.50 for working 18 hours

Monica [59]

6.75 per hour
121.50/18 = 6.75

7 0

3 years ago

Read 2 more answers

A center for medical services reported that there were 295,000 appeals for hospitalization and other services. For this group, 4

pshichka [43]

Answer:

a) 0.0025 = 0.25% probability that none of the appeals will be successful.

b) 0.0207 = 2.07% probability that exactly one of the appeals will be successful.

c) 0.9768 = 97.68% probability that at least two of the appeals will be successful

Step-by-step explanation:

For each appeal, there are only two possible outcomes. Either it is succesful, or it is not. The probability of an appeal being succesful is independent of other appeals, so we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

45% of first-round appeals were successful.

This means that $p = 0.45$

Suppose 10 first-round appeals have just been received by a Medicare appeals office.

This means that $n = 10$

(a) Compute the probability that none of the appeals will be successful.

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.45)^{0}.(0.55)^{10} = 0.0025$

0.0025 = 0.25% probability that none of the appeals will be successful.

(b) Compute the probability that exactly one of the appeals will be successful.

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{10,1}.(0.45)^{1}.(0.55)^{9} = 0.0207$

0.0207 = 2.07% probability that exactly one of the appeals will be successful.

(c) What is the probability that at least two of the appeals will be successful

This is $P(X \geq 2)$

Either less than two appeals are succesful, or at least two are. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0025 + 0.0207 = 0.0232$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0232 = 0.9768$

0.9768 = 97.68% probability that at least two of the appeals will be successful

6 0

3 years ago

Plz help with this bzbdbdbdb

forsale [732]

Answer C is correct!

3 0

3 years ago

Mike determined that some children like to eat cake, some like to eat cookies, and others don't like to eat cake or cookies. He

8_murik_8 [283]

Hey there!

This is an example of conditional probability, or P(A | B). Let's say that the probability of a child eating cake is event A, and the probability of them eating cookies is event B.

Conditional probability (A, given that B has occurred) can be represented and found by the following equation:

P(A | B) = P(A ∩ B)/P(B)

P(A ∩ B) is the probability of a child eating both cake and a cookie. This is also the middle of the venn diagram you were given. P(B) is simply the probability of event B happening, which, as we established, is eating a cookie. 

P(A | B) = P(A ∩ B) / P(B)
P(A | B) = 0.1 / 0.5
P(A | B) = 0.2

There will be a 0.2, or 20%, chance that a child will eat cake, given that they've eaten a cookie.

Sorry for the late response, but I hope this still helped you out!

5 0

3 years ago