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eduard
2 years ago
10

What is 23=5-2/3m ??????

Mathematics
1 answer:
Zielflug [23.3K]2 years ago
3 0

Answer:

Step-by-step explanation:

if it's

23 = 5 - (2/3)m

18 = (-2/3)m

m = 18(-3/2)

m = -27

if it's

23 = 5 - 2/(3m)

18 = -2/(3m)

18(3m) = -2

54m = - 2

m = -2/54

m = - 1/27

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Julie, a police officer, has the following work pattern. She works 3 days, has 4 days off, then works 4 days, then has 3 days of
zalisa [80]

Answer:

See below.

Step-by-step explanation:

For each day number, Y means working, and N means off

Day

1   Y

2   Y

3   Y

4   N

5   N

6   N

7   N

8   Y

9   Y

10   Y

11   Y

12   N

13   N

14   N

15   Y

16   Y

17   Y

18   N

19   N

20   N

21   N

22   Y

23   Y

24   Y

25   Y

26   N

27   N

28   N

29   Y

30   Y

31   Y

32   N

33   N

34   N

35   N

36   Y

37   Y

38   Y

39   Y

40   N

41   N

42   N

43   Y

44   Y

45   Y

46   N

a. Today is day 1. 30 days from today is day 31.

She works on day 31, so she does work 30 days from today.

b. 45 days from today is day 46. She is off on day 46, so she will be off 45 days from today.

8 0
3 years ago
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Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

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Answer:

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The last one isn’t
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