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frez [133]
4 years ago
8

Analyze the statement below and complete the instructions that follow. If a month is June, then it has 30 days. Write the contra

positive of the statement above. A. If a month is not June, then the month has 30 days. B. If a month does not have 30 days, then the month is June. C. If a month is June, then the month does not have 30 days. D. If a month does not have 30 days, then the month is not June.
Mathematics
1 answer:
aleksley [76]4 years ago
6 0
Note first that several months have 30 days (each):  April, June and September.

(A) could not be correct, since the month in question could have 28, 29 or 31 days.

(B) See (A), above.  If the month in question does not have 30 days, then the month could NOT be April, June or September.  Reject (B).

(C) You do this one, similarly to my responses to (A) and (B), above.

(D) This one is true, since we know that June has 30 days.  If the month in question does NOT have 30 days, then that month could not possibly be June.
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Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml
Temka [501]
<h2>Answer:</h2>

a)

The probability that both televisions work is:  0.42

b)

The probability at least one of the two televisions does not​ work is:

                          0.5833

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

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

          9-3=6

a)

The probability that both televisions work is calculated by:

=\dfrac{6_C_2}{9_C_2}

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42

b)

The probability at least one of the two televisions does not​ work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5

and the probability both do not work is:

=\dfrac{3_C_2}{9_C_2}\\\\\\=\dfrac{1}{12}\\\\\\=0.0833

Hence, Probability that atleast does not work is:

             0.5+0.0833=0.5833

4 0
4 years ago
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