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liq [111]
3 years ago
5

7) Write as a numeral: 40,000 + 7,000 + 600 + 40 + 3

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
5 0

Answer:

47,643

Step-by-step explanation:

This question is basically asking you to add up all of these numbers together and present the end result.

1) 40,000 + 7,000 = 47,000

2) 600 + 40 + 3 = 643

3) 47,000 + 643 = 47,643.

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5w−23<−3(3−4w)

Simplify both sides of the inequality.

5w−23<12w−9

Subtract 12w from both sides.

5w−23−12w<12w−9−12w

−7w−23<−9

Add 23 to both sides.

−7w−23+23<−9+23

−7w<14

Divide both sides by -7.

−7w/−7 < 14/−7

w > −2

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In February, the amount of snowfall was 2 3/8 feet. In March, the amount of snowfall was 1 1/3 feet. What was the amount of snow
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Answer:

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The general manager, marketing director, and 3 other employees of Company A are hosting a visit by the vice president and 2 othe
yawa3891 [41]

Solution :

Let the three places be 1, 2, 3, 4, 5, 6, 7, 8

a). Number of the cases when a general manager is the next to a vice president is equal to 7 and the these 2 can be arranged in 21 ways. So the total number of ways = 7 x 2

                  = 14

[(1,2)(2,1) (2,3)(3,2) (3,4)(4,3) (4,5)(5,4) (5,6)(6,5) (6,7)(7,8) (8,7)(7,6)]

Therefore the required probability is

  $=\frac{14}{8!}$

 = $\frac{14}{40320} = 0.000347$

b). The probability that the marketing director to be placed in the leftmost position is

   $=\frac{7!}{8!}$

  $=\frac{1}{8} = 0.125$

c). The two events are not independent because

   $P(A \cap B) \neq P(A) \times P(B)$

  $\frac{12}{8!} \neq \frac{14}{8!} \times \frac{1}{8}$

where A is the case a and B is the case b.

8 0
3 years ago
Am I on the right track? Or am I doing something wrong?
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It looks good to me
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3 years ago
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