The value of the 9 in 901,532 is how many times the value of the 9 in 157,900? NEED HELP!!!

B × A × C = {tall foam nonfat, tall foam whole, tall no-foam nonfat, tall no-foam whole, grande foam non fat, grande foam whole, grande no foam nonfat, grande no foam whole, venti foam nonfat,, venti foam whole, venti nofoam nonfat, venti nofoam whole}

For the element of B × C

B × C = {foam nonfat, foam whole, no foam nonfat, no foam whole}

<h3>Product of sets</h3>

Given the following sets

A = {tall, grande, venti}

B = {foam, no-foam}

C = {non-fat, whole}

The set of A * B is expressed as:

A * B = {tall foam, tall no foam, grande foam, grande no foam, venti foam, venti nofoam}

A × B × C = {tall foam nonfat, tall foam whole, tall no-foam nonfat, tall no-foam whole, grande foam non fat, grande foam whole, grande no foam nonfat, grande no foam whole, venti foam nonfat,, venti foam whole, venti nofoam nonfat, venti nofoam whole}

The element of B × A × C is equivalent to the element of A × B × C.

B × A × C = {tall foam nonfat, tall foam whole, tall no-foam nonfat, tall no-foam whole, grande foam non fat, grande foam whole, grande no foam nonfat, grande no foam whole, venti foam nonfat,, venti foam whole, venti nofoam nonfat, venti nofoam whole}

For the element of B × C

B × C = {foam nonfat, foam whole, no foam nonfat, no foam whole}

Learn more on set notation here: brainly.com/question/24462379

8 0

2 years ago

A fair coin is to be tossed 20 times. Find the probability that 10 of the tosses will fall heads and 10 will fall tails, (a) usi

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Using the distributions, it is found that there is a:

a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item a:

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

20 tosses, hence $n = 20$ .
Fair coin, hence $p = 0.5$ .

The probability is P(X = 10), thus:

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

$P(X = 10) = C_{20,10}.(0.5)^{10}.(0.5)^{10} = 0.1762$

0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item b:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.
The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item c:

For the approximation, the mean and the standard deviation are:

$\mu = np = 20(0.5) = 10$

$\sigma = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.5)} = \sqrt{5}$

Using continuity correction, this probability is $P(10 - 0.5 \leq X \leq 10 + 0.5) = P(9.5 \leq X \leq 10.5)$ , which is the p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.