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2 years ago

Simplify these expressions. a. 13xyz + 9 + 2xyz + 3 b. 6ab + (–2ab) + 4 + 6 c. 3 + 10x + (–2) + (–3x) d. 2x + 3 + x + 10

Mathematics

2 answers:

Orlov [11]2 years ago

8 0

Answer:

15 $xyz$ + 4 $ab$ + $av$ + 3 $b$ + 6 $c$ + d + 10 $x$ + 27

Step-by-step explanation:

Add The Expressions.

seraphim [82]2 years ago

5 0

Answer:

a) 15xyz +12

b) 4ab +10

c) 7x +1

d) 3x + 13

Step-by-step explanation:

Put the like terms together. Make sure you put the things with higher variable coefficiants first.

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Answer:

46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups

Step-by-step explanation:

We use two binomial trials to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

Probability of at least nine participants finishing the study in a group.

0.2 probability of a students dropping out. So 1 - 0.2 = 0.8 probability of a student finishing the study. This means that $p = 0.8$ .

10 students, so $n = 10$

We have to find:

$P(X \geq 9) = P(X = 9) + P(X = 10)$

Then

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

$P(X = 9) = C_{10,9}.(0.8)^{9}.(0.2)^{1} = 0.2684$

$P(X = 10) = C_{10,10}.(0.8)^{10}.(0.2)^{0} = 0.1074$

$P(X \geq 9) = P(X = 9) + P(X = 10) = 0.2684 + 0.1074 = 0.3758$

0.3758 probability that at least nine participants complete the study in a group.

Calculate the probability that at least nine participants complete the study in one of the two groups, but not in both groups?

0.3758 probability that at least nine participants complete the study in a group. This means that $p = 0.3758$

Two groups, so $n = 2$

We have to find P(X = 1).

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

$P(X = 2) = C_{2,1}.(0.3758)^{1}.(0.6242)^{1} = 0.4691$

46.91% probability that at least nine participants complete the study in one of the two groups, but not in both groups

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