All categories

3 years ago

12 divided by 5/6 (pretty easy but im lazy)<3

Mathematics

2 answers:

Mumz [18]3 years ago

6 0

Answer:

14.4

Step-by-step explanation:

Hope this helps you :)

Brrunno [24]3 years ago

6 0

Answer:

72/5 = 14.4

Step-by-step explanation:

You might be interested in

(you can zoom in) Can someone please explain why this is wrong?? ( Pythagorean theorem )

Serjik [45]

20^2 + 13.5^2 = c^2
582.25 = c^2
√582.25 = c
24.1 = c

hope this helps :)

6 0

3 years ago

Read 2 more answers

Suppose there was a cancer diagnostic test was 95% accurate both on those that do and those do not have the disease. If 0.4% of

Musya8 [376]

Complete Question

Suppose there was a cancer diagnostic test was 95% accurate both on those that do and 90% on those do not have the disease. If 0.4% of the population have cancer, compute the probability that a particular individual has cancer, given that the test indicates he or she has cancer.

Answer:

The probability is $P(C | A) = 0.0042$

Step-by-step explanation:

From the question we are told that

The probability that the test was accurate given that the person has cancer is

$P(A | C) = 0.95$

The probability that the test was accurate given that the person do not have cancer is

$P(A | C') = 0.90$

The probability that a person has cancer is

$P(C) = 0.004$

Generally the probability that a person do not have cancer is

$P(C') = 1- P(C)$

=> $P(C') = 1- 0.004$

=> $P(C') = 0.996$

Generally the probability that a particular individual has cancer, given that the test indicates he or she has cancer is according to Bayes's theorem evaluated as

$P(C | A) = \frac{P(A | C) * P(C)}{P(A|C) * P(C) + P(A| C') * P(C')}$

=> $P(C | A) = \frac{ 0.95 * 0.004 }{ 0.95 * 0.004 + 0.90 * 0.996}$

=> $P(C | A) = 0.0042$

7 0

3 years ago

An airplane has 100 seats for passengers. Assume that the probability that a person holding a ticket appears for the flight is 0

zmey [24]

Answer:

96.33% probability that everyone who appears for the flight will get a seat

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .