All categories

2 years ago

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:

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

You might be interested in

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 A. 2x-1=y 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 X 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 n = 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 n = 20 and p = 0.20.

The probability mass function of X 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