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drek231 [11]
2 years 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]2 years ago
7 0

Answer:

C. None of the above

Semenov [28]2 years 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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A box contains 24 transistors,4 of which are defective. If 4 are sold at random,find the following probabilities. i. Exactly 2 a
zavuch27 [327]

SOLUTION

This is a binomial probability. For i, we will apply the Binomial probability formula

i. Exactly 2 are defective

Using the formula, we have

\begin{gathered} P_x=^nC_x\left(p^x\right?\left(q^{n-x}\right) \\ Where\text{ } \\ P_x=binomial\text{ probability} \\ x=number\text{ of times for a specific outcome with n trials =2} \\ p=\text{ probability of success = }\frac{4}{24}=\frac{1}{6} \\ q=probability\text{ of failure =1-}\frac{1}{6}=\frac{5}{6} \\ ^nC_x=\text{ number of combinations = }^4C_2 \\ n=\text{ number of trials = 4} \end{gathered}

Note that I made the probability of being defective as the probability of success = p

and probability of none defective as probability of failure = q

Exactly 2 are defective becomes the binomial probability

\begin{gathered} P_x=^4C_2\times\lparen\frac{1}{6})^2\times\lparen\frac{5}{6})^{4-2} \\ P_x=6\times\frac{1}{36}\times\frac{25}{36} \\ P_x=\frac{25}{216} \\ =0.1157 \end{gathered}

Hence the answer is 0.1157

(ii) None is defective becomes

\begin{gathered} \lparen\frac{5}{6})^4=\frac{625}{1296} \\ =0.4823 \end{gathered}

hence the answer is 0.4823

(iii) All are defective

\begin{gathered} \lparen\frac{1}{6})^4=\frac{1}{1296} \\ =0.00077 \end{gathered}

(iv) At least one is defective

This is 1 - probability that none is defective

\begin{gathered} 1-\lparen\frac{5}{6})^4 \\ =1-\frac{625}{1296} \\ =\frac{671}{1296} \\ =0.5177 \end{gathered}

Hence the answer is 0.5177

3 0
1 year ago
What is the value of the exponential function f (x)=2 times (1/3)^x when x= -2
frutty [35]

Answer:

18

Step-by-step explanation:

f(x) = 2(1/3)^x

[x = -2]

f(-2) = 2(1/3)⁻²

= 2(9)

= 18

5 0
2 years ago
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