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drek231 [11]
1 year ago
11

Which expressions are equivalent to (5g+3h+4)\cdot2(5g+3h+4)⋅2left parenthesis, 5, g, plus, 3, h, plus, 4, right parenthesis, do

t, 2 ?
Choose all answers that apply:

Choose all answers that apply
A
(5g+3h)\cdot8(5g+3h)⋅8left parenthesis, 5, g, plus, 3, h, right parenthesis, dot, 8

(Choice B)
B
(5g+3h)\cdot6(5g+3h)⋅6left parenthesis, 5, g, plus, 3, h, right parenthesis, dot, 6

(Choice C)
C
None of the above
Mathematics
2 answers:
lions [1.4K]1 year ago
7 0

Answer:

C. None of the above

Semenov [28]1 year ago
6 0

Answer:

C None of the above

Step-by-step explanation:

The expression

(5g+3h+4)⋅2

can be expanded using distributive property as follows:

5g⋅2 + 3h⋅2 + 4⋅2  =

= 10g + 6h + 8

option A expression

(5g+3h)⋅8

can be expanded using distributive property as follows:

5g⋅8+3h⋅8 =

= 40g + 24h

which is different from 10g + 6h + 8

Option B expression

(5g+3h)⋅6

can be expanded using distributive property as follows:

5g⋅6+3h⋅6 =

= 30g + 18h

which is different from 10g + 6h + 8

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Answer:

(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

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Step-by-step explanation:

The complete question is:

A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is

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Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.

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P(X)=p=\frac{10}{18}=0.556

A randomly selected egg is unspoiled or not is independent of the others.

It is provided that a chef picks 5 eggs at random.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.556.

The success is defined as the selection of an unspoiled egg.

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

P(X=x)={5\choose x}(0.556)^{x}(1-0.556)^{5-x};\ x=0,1,2,3...

(a)

Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:

P(X=5)={5\choose 5}(0.556)^{5}(1-0.556)^{5-5}\\=1\times 0.05313\times 1\\=0.0531

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(b)

Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

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Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c)

Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:

P (X > 1) = 1 - P (X ≤ 1)

              = 1 - P (X = 0) - P (X = 1)

              =1-\sum\limits^{1}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=1-0.0173-0.1080\\=0.8747

Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

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