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3 years ago

Which comparison is not correct

Mathematics

2 answers:

leva [86]3 years ago

5 0

|8|>|-9| is going to be the one that is not correct

gtnhenbr [62]3 years ago

4 0

Answer:

A, The first comparison is not correct.

Step-by-step explanation:

When you have -|2| (example) then it would equal -2

When you have |-2| (example) then it would equal 2

here it is saying |-2| < |-9|

It is basically saying -2 < -9

-9 is not greater than -2 therefore the first option is an incorrect comparison.

therefire making it the correct answer... I hope this helps. please mark brainliest.

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

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

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

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

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(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

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

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