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matrenka [14]
3 years ago
11

50,000 contestants participate in an on-line game. The game randomly eliminates 20% of the contestants each day. Determine the n

umber of contestants remaining in the game after one week.
Mathematics
1 answer:
irga5000 [103]3 years ago
6 0

Answer:10,486 contestants

Step-by-step explanation:

First day:

Total number of contestants at the beginning = 50,000

After first day, 20% 0f 50,000 are removed.

That is, (20/100)×50,000 = 10,000

Contestants remaining after first day = 50,000-10,000 = 40,000

Second day:

Contestants= 40,000

20% removed

That is (20/100)×40000=8000

Balance =40000-8000=32,000

Third day:

Contestants= 32000

20% removed

That is (20/100)×32000=6400

Balance =32000-6400=25,600

Fouth day:

Contestants= 25,600

20% removed

That is (20/100)×25,600=5,120

Balance =25,600-5,120=20,480

Fifth day:

Contestants= 20,480

20% removed

That is (20/100)×20,480=4096

Balance =20,480-4096=16,384

Sixth day:

Contestants= 16,384

20% removed

That is (20/100)×16,384=3276.8

Balance =16,384-3276.8=13,107.2

Seventh day:

Contestants= 13,107.2

20% removed

That is (20/100)×13,107.2=2621.44

Balance =13107.2-2621.44=10485.76

Therefore, number of contestants remaining after one week equals 10,486.

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16 squared with explanation.
pochemuha

Answer:

256

Step-by-step explanation:

Multiply 16 by 16.

You bet 256, right?

The answer is 256 now.

Hope it helps!

Formula: multiply number you have x number you have.

7 0
3 years ago
Read 2 more answers
Nuts.com sells raw almonds for $10 per pound and roasted pistachios for $12 per pound. Christopher spends $70, before taxes and
galben [10]

Answer:

second equation

Step-by-step explanation:

10a + 12p = 70  (10 goes with almonds [a], 12 goes with pistachios[p], $10 and $12 is price per pound and $70 is total price, so that's why its = to 70)

a + p = 6.3        (this is total almonds and pistachios adding up to 6.3 pounds)

a = pounds of almonds

p = pounds of pistachios

3 0
3 years ago
Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
3 years ago
Beth is going to enclose a rectangular area in back of her house. the house wall will form one of the four sides of the fenced i
Alina [70]

The maximum possible area would have a length of 24 feet and width of 12 feet.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

An independent variable is a variable that does not depend on other variables while a dependent variable is a variable that depends on other variables.

Let x represent the length and y represent the width, hence:

Since beth has 48 ft fencing and cover 3 sides, hence:

x + 2y = 48  

x = 48 - 2y     (1)

Also:

Area (A) = xy

A = (48 - 2y)y

A = 48y - 2y²

The maximum area is at A' = 0, hence:

A' = 48 - 4y

48 - 4y = 0

y = 12 feet

x = 48 - 2(12) = 24

The maximum possible area would have a length of 24 feet and width of 12 feet.

Find out more on equation at: brainly.com/question/2972832

#SPJ1

7 0
2 years ago
4,094 divided by 46 please help
Arlecino [84]
The answer to your question is 89
3 0
3 years ago
Read 2 more answers
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