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Sonja [21]
2 years ago
11

Please answer this question correctly and i'll mark you as brainlilest! Thank you.

Mathematics
1 answer:
raketka [301]2 years ago
4 0

Answer:

\frac{ {7}^{4} ( - 2) {}^{4} {11}^{3}  6}{7 {}^{3}    {2}^{2} {6}^{2} 11 {}^{2}   }  =  \frac{7 \times  {2}^{2}  \times 11}{6}  =  \frac{308}{6}  =  \frac{154}{3}   \\  \\  \frac{ {7}^{0} {12}^{0}  {17}^{0}  }{ {57}^{0}  + ( {42}^{0}  \times ( {13}^{0} ))}  =  \frac{1}{1 + 1}  =  \frac{ 1}{2}

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PLEASE HELP 10 POINTS and please explain -32.48-(14.014)
Ostrovityanka [42]

-32.48-(14.014)

- (32.48 + 14.014)

add 32.48 +14.014 by lining up the decimal

 32.48

+ 14.014

------------

46.494

then bring back the negative

-(46.494)

Answer: -46.494

5 0
3 years ago
A and B are sumplimenteey . m B =121 find m A
vichka [17]

Answer:

m∠A = 59°

Step-by-step explanation:

Supplementary angles mean that, when the two are combined, their total measurement will equal 180°.

It is given that one of the angles is 121°. Subtract 121 from 180:

180 - 121 = 59

m∠A = 59°

~

4 0
2 years ago
Explain how to graph a line from the following equation <br><br> A. 2x-1=y<br> B. Y=-9/2x-2/8
antiseptic1488 [7]

Answer:  its b

Step-by-step explanation:

4 0
2 years ago
Two athletic teams play a series of games; the first team to win 4 games is declared the overall winner. Suppose that one of the
kodGreya [7K]

Answer:

  • p=0.7103 (4-game series)
  • p=0.6480 (2-game series)

Step-by-step explanation:

Let X be the random variable equal the the first 4 straight wins. An overall win for the stronger team implies a negative binomial function with the parameters  n=4, p=0.6:

P(X=4)={{i-1}\choose {4-1}}0.6^40.4^{i-4},\  i=4,5,6,7

#We find probabilities for the different values of i:

P(X=4)={3\choose 3}0.6^4=0.1296\\\\P(X=5)={4\choose 3}0.6^40.4^1=0.2074\\\\P(X=6)={5\choose 3}0.6^40.4^2=0.2074\\\\P(X=4)={6\choose 3}0.6^40.4^3=0.1659

Hence, probability of the stronger team winning overall is:

=P(X=4)+P(X=5)+P(X=6)+P(X=7)\\\\=0.7103

#Define Y as the random variable for winning 2/3 games.:

P(Y=2)={1\choose 1}0.6^2=0.3600\\\\P(Y=3)={2\choose3}0.6^20.4=0.2880\\\\P(win)=0.2880+0.3600=0.6480

Hence, probability of the stronger team winning in 2 out 3 game series is 0.6480

The stronger team has a higher chance of winning in a 4-game series(0.7103>0.6480)

8 0
2 years ago
A multiple-choice examination has 20 questions, each with five possible answers, only one of which is correct. Suppose that one
kari74 [83]

Answer:

The probability that the student answers at least seventeen questions correctly is 8.03\times 10^{-10}.

Step-by-step explanation:

Let the random variable <em>X</em> represent the number of correctly answered questions.

It is provided all the questions have five options with only one correct option.

Then the probability of selecting the correct option is,

P(X)=p=\frac{1}{5}=0.20

There are <em>n</em> = 20 question in the exam.

It is also provided that a student taking the examination answers each of the questions with an independent random guess.

Then the random variable can be modeled by the Binomial distribution with parameters <em>n</em> = 20 and <em>p</em> = 0.20.

The probability mass function of <em>X</em> is:

P(X=x)={20\choose x}\ 0.20^{x}\ (1-0.20)^{20-x};\ x =0,1,2,3...

Compute the probability that the student answers at least seventeen questions correctly as follows:

P(X\geq 17)=P (X=17)+P (X=18)+P (X=19)+P (X=20)

=\sum\limits^{20}_{x=17}{{20\choose x}\ 0.20^{x}\ (1-0.20)^{20-x}}\\\\=0.00000000077+0.000000000032+0.00000000000084+0.000000000000042\\\\=0.000000000802882\\\\=8.03\times10^{-10}

Thus, the probability that the student answers at least seventeen questions correctly is 8.03\times 10^{-10}.

4 0
3 years ago
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