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3 years ago

Find the midpoint between A and C.

Mathematics

2 answers:

Mariana [72]3 years ago

4 0

Answer:

can't see the answer choices correctly they are blury

Step-by-step explanation:

Evgen [1.6K]3 years ago

4 0

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{(0. \: 5, \: 0.5 \: )}}}}}$

Step-by-step explanation:

Given,

A ( -2 , 4 )⇒ ( x₁ , y₁ )

C ( 3 , - 3 )⇒ ( x₂ , y₂ )

Finding the midpoint between A and C

$\boxed{ \sf{midpoint = ( \frac{x1 + x2}{2} \: , \: \frac{y1 + y2}{2} )}}$

$\dashrightarrow{ \sf{( \frac{ - 2 + 3}{2} \: , \: \frac{4 + ( - 3)}{2} )}}$

$\dashrightarrow{ \sf{( \frac{1}{2} \:, \: \frac{4 - 3}{2} )}}$

$\dashrightarrow{ \sf{( \frac{1}{2} \: , \: \frac{1}{2} }})$

$\dashrightarrow{( \sf{0.5 \: , \: 0.5) \: }}$

Hope I helped!

Best regards! :D

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Following are the solution to the given question:

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3 years ago

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zavuch27 [327]

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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

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

Hence the answer is 0.1157

(ii) None is defective becomes