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gulaghasi [49]
3 years ago
8

A toy manufacturer inspects boxes of toys before shipment. Each box contains 7 toys. The inspection procedure consists of random

ly selecting three toys from the box. If one or more of the toys are defective, the box is not shipped. Suppose that a given box has two defective toys. What is the probability that it will be shipped?
a) 0.0486
b) 0.2857
c) 0.1429
d) 0.7143
e) 0.3088"
Mathematics
1 answer:
leva [86]3 years ago
3 0

Answer:

B-) 0.2857

Step-by-step explanation:

So to be shipped all of the toys, toys need to be not faulty. This means that we will think about the undefective ones.

We have :

2 defective toys

5 undefective toys

To be shipped all of the toys must be undefective.

We will have to choose undefective ones from all the toys, 3 times:

Undefective ones      Undefective ones     Undefective ones

--------------------------- . ----------------------------- .  --------------------------

      All the toys                 All the toys                 All the toys

> 5/7 . 4/6 . 3/5

><em>0.2857</em>1428571

So, the answer is B-) 0.2857

<em>I hope it will be understood.</em>

<em>If I have any inaccuracies please let me know.</em>

<em>Have a nice day and never stop questioning!</em>

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The function y - 3 = 4(x + 2) is in point-slope form. Select the equivalent equation in slope-intercept form
Phoenix [80]
<h2>Answer:</h2>

All we need to do is solve for y.

y - 3 = 4(x + 2)\\\\y - 3 = 4x + 8\\\\y = 4x + 11

The answer is <em>A. y = 4x + 11</em>.

7 0
3 years ago
PLZZZZZZZZZZ HELP
MrRissso [65]
<h2>Answer:</h2>

The correct answer is:   Option: A

A.   85, 78, 80, 108, 46, 66, 68, 82, 72, 68

<h2>Step-by-step explanation:</h2>

It is given that:

The whisker ranges from 66 to 85.

The box ranges from 68 to 82 with the vertical bar inside the box at 75.

  • Hence, the minimum value of the box plot is: 66
  • Maximum value of box plot is: 85

Also,

  • The first quartile or the lower quartile i.e. Q_1 is: 68
  • The middle quartile or median i.e. Q_2 is: 75
  • The upper quartile or third quartile i.e. Q_3 is: 82

Also, it is given that:

One dot mark above forty-six.

One dot mark above one hundred and eight.

A)

On arranging the data in increasing order after removing outliers 46 and 108 we have:

 66    68    68    72    78    80    82     85  

Hence, from this data we have:

Minimum value=66

maximum value=85

Q_1=68\\\\Q_2=75\\\\\\Q_3=81

As all the values matches the value defined.

Hence, this option is correct.

B)

On arranging the data in increasing order after removing 46 and 108 we have:

66    68    68    70   70   80    82     85

Hence, from this data we have:

Minimum value=66

maximum value=85

Q_1=68\\\\Q_2=70\\\\\\Q_3=81

As the value of Q_2=70\neq 75.

Hence, this option is incorrect.

C)

On arranging the data in increasing order after removing 46 and 108 we have:

66    68    72    78   80   80    82     85

Hence, from this data we have:

Minimum value=66

maximum value=85

Q_1=70\\\\Q_2=79\\\\\\Q_3=81

As the value of  Q_1,Q_2\ and Q_3 do not match.

Hence, this option is incorrect.

D)

On arranging the data in increasing order after removing 46 and 108 we have:

66    68    68    72   78   80    80     85

Hence, from this data we have:

Minimum value=66

maximum value=85

Q_1=68\\\\Q_2=79\\\\\\Q_3=80

As the value of  Q_3=80\neq 82 do not match.

Hence, this option is incorrect.

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3 years ago
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Answer:

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Step-by-step explanation:

8 0
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Factor: 2x^2 - 4x + 1
shepuryov [24]
Yes that’s it son welp ok
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The student council is planning an activity night to raise money. The student council will charge $5 per student to attend the a
sineoko [7]

Answer:

5n

Step-by-step explanation:

They are charging 5 dollars per person and n is the number of students.  You would multiply 5 by n (the number of students) to see how much money they raised.

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