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Zigmanuir [339]
3 years ago
9

Anyone know how to solve this?

Mathematics
1 answer:
yuradex [85]3 years ago
5 0

Answer:

<h2>\frac{16}{25}</h2>

Step-by-step explanation:

<h3>( -  \frac{4}{5})^{2}</h3><h3>{( - 1)}^{2}  \frac{ {4}^{2} }{ {5}^{2} }</h3><h3>1 \frac{ {4}^{2} }{ {5}^{2} }</h3><h3>\frac{ {4}^{2} }{ {5}^{2} }</h3><h3>\frac{16}{25}</h3><h3>Hope it is helpful....</h3>
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Suppose we want to choose 5 objects, without replacement, from 15 distinct objects.
goblinko [34]

15444 ways we can choose 5 objects, without replacement, from 15 distinct objects.

Given that, suppose we want to choose 5 objects, without replacement, from 15 distinct objects.

<h3>What is a permutation?</h3>

A permutation is a mathematical calculation of the number of ways a particular set can be arranged, where the order of the arrangement matters.

Now, 13P_{5}= 13!/(13-5)!

= 13!/8! = 13x12x11x10x9= 1287 x 120 = 15,444

Therefore, 15444 ways we can choose 5 objects, without replacement, from 15 distinct objects.

To learn more about the permutation visit:

brainly.com/question/1216161.

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3 0
1 year ago
2
Nesterboy [21]

Answer:

did you got the answer

Step-by-step explanation:

because I need it too

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I've checked out and its AAA does not prove congruence. Triangles are not congruent. They are similar

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2 years ago
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Which product is negative​
vovangra [49]

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4th option because the two negatives will become positive and the 3rd will make it negative again

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Pythagoras Theorem and Quadratic Equations<br> Solve for x:<br> a= (x-2)<br> b= 9<br> c= (x+1)
Alex_Xolod [135]

Answer:

Step-by-step explanation:

Pythagoras theorem

a^2=b^2+c^2

(x-2)^2=9^2+(x+1)^2

x^2-4x+4=81+x^2+2x+1

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-4x-2x=82-4

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The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%
liq [111]
<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

   Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

   From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by \frac{100+1.8}{100} = \frac{101.8}{100}.

   So, the population in the year t can be given by P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}

   Population in the year 2000 = 3,381,000\textrm{x}(\frac{101.8}{100})^{6}=3,762,979.38

Population in year 2000 = 3,762,979

   Let us assume population doubles by year y.

2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}

log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})

y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537

y≈2033

∴ By 2033, the population doubles.

4 0
3 years ago
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