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3 years ago

Math problem, need help

Mathematics

1 answer:

vazorg [7]3 years ago

6 0

Answer:

(7-3i)

Step-by-step explanation:

You add them together so 3+4 and 2i + - 5i but since the 5 is negative it would be negaitive

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The initial population of a town is 2500, and it grows with a doubling time of 10 years. What will the population be in 8 years?

alisha [4.7K]

2500x0.80=2000 so i think it would be 2000 in 8 years

or if youre adding the 2000 to the already existing 2500 then it would be 4500.

4 0

3 years ago

EXERCISE

Olin [163]

Answer:

1)The probability that the ball drawn is not red is 6 /14

2)Probability of getting either a prime or a multiple of 3 is 7/10

3)Probability of drawing three balls without replacement such that all three are Same colour is 0.0747

Probability of drawing three balls without replacement such that all three are different colour is 0.5

Step-by-step explanation:

1) A ball is drawn randomly from a bag containing 8 red, 4 blue and 2 green identical balls. Find the probability that the ball drawn is not red.

No. of red balls = 8

No. of blue balls = 4

No. of green balls = 2

Total no. of balls = 8+4+2=14

P(Getting red bal)= $\frac{8}{14}$

P( ball drawn is not red) = $1 - \frac{8}{14}=\frac{6}{14}$

So,the probability that the ball drawn is not red is 6 /14

2)What is the probability that a number chosen at random from the integers between 1 and 10 inclusive is either a prime or a multiple of 3?

Numbers : 1 ,2,3,4,5,6,7,8,9,10

Prime numbers: 2,3,5,7 = 4

Multiple of 3 = 3,6,9 =3

So, Probability of getting prime number = $\frac{4}{10}$

Probability of getting multiple of 3 = $\frac{3}{10}$

So, Probability of getting either a prime or a multiple of 3= $\frac{4}{10}+\frac{3}{10}=\frac{7}{10}$

Hence Probability of getting either a prime or a multiple of 3 is 7/10

3)A bag contains four red balls, five blue balls and six green balls. Three balls are drawn from the bag at random without replacement. What is the probability that all three are:

The same colour

Different colour

No. of red balls = 4

No. of blue balls = 5

No. of green balls = 6

Total balls = 15

Case 1 :all three are Same colour

So, probability of drawing three balls without replacement such that all three are Same colour = $\frac{4}{15} \times \frac{3}{14} \times \frac{2}{13}+\frac{5}{15} \times \frac{4}{14} \times \frac{3}{13}+\frac{6}{15} \times \frac{5}{14} \times \frac{4}{13}=0.0747$

So, probability of drawing three balls without replacement such that all three are different colour = $\frac{4}{10} \times \frac{5}{9} \times \frac{6}{8}+\frac{5}{10} \times \frac{4}{9} \times \frac{6}{8}+\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}=0.5$