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2 years ago

7

0.989, 1.118, 1.637, 1.83, 1.01, 1.912, 1.115, 2.504, 2.2,1 3.3, 1.127, 1.281, 2.4, 0.356, 653.63, 1.714, 2.036, 1.305 from leas

t to greatest

2 answers:

Norma-Jean [14]2 years ago

8 0

Answer:

0.356, 0.989, 1.01, 1.115, 1.118, 1.127, 1.281, 1.305, 1.637, 1.714, 1.83, 1.912, 2.036, 2.2, 2.4, 2.504, 13.3, and 653.63

Alenkinab [10]2 years ago

6 0

0.356/ 0.989/ 1.01/ 1.115/1.118/1.127/1.281/1.305/1.637/1.714/1.83/1.912/2.036/2.21/2.4/2.504/3.3

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X+y+z=−4<br> 2x+3y−2z=10<br> −x+2y−3z=−12 <br> solve for x,y,z<br> WILL MARK BRAINLIEST!!!

Norma-Jean [14]

Answer:

x = 13

y = -10

z = -7

Step-by-step explanation:

x+y+z=−4

2x+3y−2z=10

−x+2y−3z=−12

Add the first and the third equations to eliminate x

x+y+z=−4

−x+2y−3z=−12

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

3y -2z = -16 Equation A

Add the second and twice the third equation to eliminate x

2x+3y−2z=10

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3 years ago

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2 years ago

Suppose that an experiment consists of flipping a coin 3 times and observing the resulting sequence of heads and tails. Find the

ki77a [65]

Using the binomial distribution, it is found that there is a 0.125 = 12.5% probability of observing exactly 3 tails.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem, considering 3 tosses of a fair coin, the parameters are n = 3 and p = 0.5.

The probability of 3 tails is P(X = 3), hence:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{3,3}.(0.5)^{3}.(0.5)^{0} = 0.125$

0.125 = 12.5% probability of observing exactly 3 tails.

More can be learned about the binomial distribution at brainly.com/question/24863377

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