93 times 34, I'm multiplying whole numbers.

A container contains 12 diesel engines. The company chooses 5 engines at random, and will not ship the container if any of the

Rufina [12.5K]

Answer: $\frac{^{10}C_5}{^{12}C_5}$

Step-by-step explanation:

Given a total of 12 diesel engines

out of which 2 are defective

Company have to choose 5 good engines to ship.

Therefore no of ways in good engines are selected is $^{10}C_5$

no of ways in which any 5 engines are selected from a total of 12 engines is

$^{12}C_5$

Therefore the required probability = $\frac{^{10}C_5}{^{12}C_5}$

= $\frac{7}{22}$ =0.318

3 0

3 years ago

A store sells 12-packs of ballpoint pens for $1.09 each and composition books for $2.79 each. Cindy buys m packs of ballpoint pe

Yuri [45]

Answer: im not very sure and im sorry but the answer is either D or C

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A practice law exam has 100 questions, each with 5 possible choices. A student took the exam and received 13 out of 100.If the s

Cloud [144]

Answer:

$z=\frac{13-20}{4}=-1.75$

Assuming:

H0: $\mu \geq 20$

H1: $\mu$

$p_v = P(Z$

Step-by-step explanation:

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".

Let X the random variable of interest (number of correct answers in the test), on this case we now that:

$X \sim Binom(n=100, p=0.2)$

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:

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

We need to check the conditions in order to use the normal approximation.

$np=100*0.2=20 \geq 10$

$n(1-p)=20*(1-0.2)=16 \geq 10$

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

$E(X)=np=100*0.2=20$

$\sigma=\sqrt{np(1-p)}=\sqrt{100*0.2(1-0.2)}=4$

So we can approximate the random variable X like this:

$X\sim N(\mu =20, \sigma=4)$

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean". The letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z$