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sineoko [7]
2 years ago
14

Calculate the approximate value of the 82nd percentile. Enter an exact value.

Mathematics
1 answer:
irinina [24]2 years ago
5 0
I need more information and data for the question in order for me to answer it.
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Can anyone help me out? I’ll give Brainly! Thank you :)
Oliga [24]

Answer:

The rate of return is 14%

Step-by-step explanation:

The rate of return can be determined by,

RR = \frac{A_{f} - A_{i}  }{A_{i} } x 100%

where:

RR is the rate of return

A_f} is the final amount = $690 -  $6 = $684

A_{i} is the initial amount = $15 x 40 = $ 600

So that,

RR = \frac{684 - 600}{600} x 100%

     = 0.14 x 100%

     = 14%

Therefore, the rate of return is 14%.

7 0
2 years ago
Please answer the following questions with the proper significant figures:
mamaluj [8]

Answer:

idont know what's the answer but goodluck bro<em><u> </u></em><em><u>loveyou</u></em>

5 0
3 years ago
Domain and range and function , need ASAP
Diano4ka-milaya [45]

Answer:

Domain: -2\leq x\leq 2

Range: 0\leq y\leq 4

Step-by-step explanation:

The domain of a function is the set of values that one can input into a function and get a valid result.

The range of a function is the set of valid outputs that one can attain when a value is substituted into a function.

6 0
3 years ago
A polynomial has a leading coefficient of 1 and the
avanturin [10]

Answer: it’s C and the second part is C as well

Step-by-step explanation: Just did it on edg! :))

3 0
2 years ago
In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any
Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

E \cap F = \emptyset,\quad E \cup F = \Omega

where \Omega is the whole sample space.

This implies that

P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day d_1.

The second footballer can be born on any other day, so he has 364 possible birthdays:

d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}

the probability for the first two footballers to be born on two different days is thus

1 \cdot \dfrac{364}{365} = \dfrac{364}{365}

Similarly, the third footballer can be born on any day, except d_1 and d_2:

d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}

so, the probability for the first three footballers to be born on three different days is

1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

So, the probability that at least two footballers are born on the same day is

1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

since the two events are complementary.

8 0
3 years ago
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