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Artyom0805 [142]

3 years ago

13

if you are the first one to answer this question, you will be marked brsinlyest (ik I spelled that wrong but yk what I mean) wha

t is 20+20x17-12 use mdas (my dear aunt sally) to answer.

1 answer:

Nina [5.8K]3 years ago

4 0

Answer:

348

Step-by-step explanation:

Multiply 20 x 17 first. Then add that value to 20 and -17. The result is the answer.

I get 348

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Select the postulate that is illustrated for the real numbers.

Paladinen [302]

<span>25 + 0 = 25

</span><span>Additive identity</span>

4 0

3 years ago

Read 2 more answers

Suppose n people, n ≥ 3, play "odd person out" to decide who will buy the next round of refreshments. The n people each flip a f

blondinia [14]

Answer:

Assume that all the coins involved here are fair coins.

a) Probability of finding the "odd" person in one round: $\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}$ .

b) Probability of finding the "odd" person in the $k$ th round: $\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left( 1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}$ .

c) Expected number of rounds: $\displaystyle \frac{2^{n - 1}}{n}$ .

Step-by-step explanation:

<h3>a)</h3>

To decide the "odd" person, either of the following must happen:

There are $(n - 1)$ heads and $1$ tail, or
There are $1$ head and $(n - 1)$ tails.

Assume that the coins here all are all fair. In other words, each has a $50\,\%$ chance of landing on the head and a

The binomial distribution can model the outcome of $n$ coin-tosses. The chance of getting $x$ heads out of

The chance of getting $(n - 1)$ heads (and consequently, $1$ tail) would be $\displaystyle {n \choose n - 1}\cdot \left(\frac{1}{2}\right)^{n - 1} \cdot \left(\frac{1}{2}\right)^{n - (n - 1)} = n\cdot \left(\frac{1}{2}\right)^n$ .
The chance of getting $1$ heads (and consequently, $(n - 1)$ tails) would be $\displaystyle {n \choose 1}\cdot \left(\frac{1}{2}\right)^{1} \cdot \left(\frac{1}{2}\right)^{n - 1} = n\cdot \left(\frac{1}{2}\right)^n$ .

These two events are mutually-exclusive. $\displaystyle n\cdot \left(\frac{1}{2}\right)^n + n\cdot \left(\frac{1}{2}\right)^n = 2\,n \cdot \left(\frac{1}{2}\right)^n = n \cdot \left(\frac{1}{2}\right)^{n - 1}$ would be the chance that either of them will occur. That's the same as the chance of determining the "odd" person in one round.

<h3>b)</h3>

Since the coins here are all fair, the chance of determining the "odd" person would be $\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}$ in all rounds.

When the chance $p$ of getting a success in each round is the same, the geometric distribution would give the probability of getting the first success (that is, to find the "odd" person) in the $k$ th round: $(1 - p)^{k - 1} \cdot p$ . That's the same as the probability of getting one success after $(k - 1)$ unsuccessful attempts.

In this case, $\displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}$ . Therefore, the probability of succeeding on round $k$ round would be

$\displaystyle \underbrace{\left(1 - n \cdot \left(\frac{1}{2}\right)^{n - 1}\right)^{k - 1}}_{(1 - p)^{k - 1}} \cdot \underbrace{n \cdot \left(\frac{1}{2}\right)^{n - 1}}_{p}$ .

<h3>c)</h3>

Let $p$ is the chance of success on each round in a geometric distribution. The expected value of that distribution would be $\displaystyle \frac{1}{p}$ .

In this case, since $\displaystyle p = n \cdot \left(\frac{1}{2}\right)^{n - 1}$ , the expected value would be $\displaystyle \frac{1}{p} = \frac{1}{\displaystyle n \cdot \left(\frac{1}{2}\right)^{n - 1}}= \frac{2^{n - 1}}{n}$ .

7 0

3 years ago

A random sample of 10 subjects have weights with a standard deviation of kg. What is the variance of their weights? be sure to

Vera_Pavlovna [14]

The variance of their weights is 128.87270 kg².

It is required to find the variance of their weights.

<h3>What is standard deviation ?</h3>

A standard deviation (or σ) is a measure of how dispersed the data is in relation to the mean. Low standard deviation means data are clustered around the mean, and high standard deviation indicates data are more spread out.

Given:

Let n=sample size=10 subjects

Standard deviation=11.352211.3522 kg

Using this formula

Variance=(Standard deviation)²

By the formula we get,

Variance=(11.352211.3522 kg)²

Variance=128.87270 kg²

Therefore, the variance of their weights is 128.87270 kg².

Learn more about standard deviation here:

brainly.com/question/16555520

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7 0

2 years ago

Mandy obtains a $155,000 20/6 balloon mortgage with a rate of 4.25%. What will her monthly payments be?

elena55 [62]

Answer:

The answer to the question is

Her monthly payments will be $279.98

Step-by-step explanation:

To solve the question we apply the equation for monthly payments as follows

Fixed monthly mortgage repayment formula is

$A = P*r*\frac{r(1+r)^n}{(1+r)^n -1}$

Where

A = Payment amount per month

r = interest rate = 4.25 %

n = number of months = 20×12 months or 240 months

P = Mortgage value = $155,000

Therefore A = $(0.0425)(155000)\frac{0.0425(1+0.0425)^{240}}{(1+0.0425)^{240} -1}$ = $279.98

For balloon mortgage the loan balance is paid off or refinanced at the end of the loan term

4 0

3 years ago

Solve 7b 3 - 4b = 3 - 3 (b 4)

frez [133]

7b + 3 - 4b = 3 - 3(b + 4)
3b + 3 = 3 - 3b - 12
3b + 3b = 3 - 12 - 3 = -12
6b = -12
b = -12/6 = -2
b = -2

8 0

3 years ago