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3 years ago

9f + 4 - 7f = 8 pls help

Mathematics

2 answers:

viva [34]3 years ago

6 0

answer is 2ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

Yuri [45]3 years ago

4 0

Step-by-step explanation:

f = 2 is the correct answer

plz mark my answer as brainlist plzzzz vote me also and hope this helps you ...

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Describe and correct the error in the solution, make sure to show your work

ASHA 777 [7]

Hope i helped!!! good luck.

5 0

3 years ago

A few weeks into the deadly SARS (Severe Acute Respiratory Syndrome) epidemic in 2003, the number of cases was increasing by abo

DaniilM [7]

Answer: The required exponential model is $1804(1.04)^t$

Step-by-step explanation:

Since we have given that

Initial cases = 1804

Rate of growth = 4%

Since it is increasing exponentially,

So, it becomes,

$A=ab^t$

Here, a = 1804

b = $1+r=1+0.04=1.04$

So, it becomes,

$A=1804(1.04)^t$

Hence, the required exponential model is $1804(1.04)^t$

6 0

3 years ago

SOMEONE PLEASE HELP ME ASAP PLEASE!!!

Strike441 [17]

Answer:

105.12 ft^2

Step-by-step explanation:

Area of a rectangle: bh

In this case 8*10.... so area of the rectangle is 80

Area of a circle: pir^2

Half it for a semicircle.

so 1/2 pi r^2

radius is 4 cuz its half of 8.

so 1/2(3.14)(4^2)=(0.5)(3.14)(16)=25.12

Now add up 80+25.12

Total is 105.12

Hope I helped :)

5 0

3 years ago

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:

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.

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

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

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

I need help with this part on the quiz

dedylja [7]

Non linear and increasing

4 0

3 years ago

Read 2 more answers