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2 years ago

A city of 100,000 is having pollution problems and is decreasing in size 2% annually (every year). Find

Mathematics

1 answer:

sweet [91]2 years ago

4 0

Answer:

13,262.

Step-by-step explanation:

The town will reduce at 2% on reducing balance method of end of each year population.

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Help plz!!!!!?!!!?!??!????

aleksandrvk [35]

700ml=23.6698fl
64-23.6698=40.3302
40.3302fl

5 0

2 years ago

Is 500 x 30 a good estimate for the number of tickets sold at the number of tickets sold at the theater in one month

salantis [7]

Answer:

It depends if business is steady. If they make about 500 a day that is a fair estimate. Make sure that the month you are talking about has 30 days.

3 0

2 years ago

last week a worm was 10 millimeters long. This week it is 13 millimeters long. What is the percent of increase of the worm's len

Sladkaya [172]

Answer:

23% increase

Step-by-step explanation:

13 - 10 = 3

3/13 = 0.2307 = 23%

Hope that helped!!! k

7 0

2 years ago

Read 2 more answers

If a 10 symbol sequence is sent through the channel,what is the probability that up to 3 symbols are in error out of the 10trans

lina2011 [118]

Answer: 0.171887

Step-by-step explanation:

Given that S0 and S1 are binary symbol of equal probabilty;

P(S0) = P(S1) = 0.5

This probability is a Binomial random variable of sequence Sn, where Sn counting the number of success in a repeated trials.

P(Sn =X) = nCx p^x (1-p)^(n-1)

Pr(at most 3) = P(0<= x <=3) = P(X=0) + 0) + P(X=1) + P(X=2) + P(X=3)

Since there are only 2 values that occur in sequence 0 and 1 ( or S0 and S1).Let the distribution be given by the sequence (0111111111),(1011111111),(11011111111),...(1111111110) for Sn= 1 is the sequence for 1 error.

10C0, 10C1, 10C2, and 10C3 is the number of sequences in value for X= 0, 1, 2, 3 having value 0 and others are 1. Let the success be p(S0)=0.5 and p(S1)= 0.5

P(0<= X <=3) = 10C0 × (0.5)^0 × (0.5)^10 + 10C1× (0.5)¹ × (0.5)^9 + 10C2 × (0.5)² × (0.5)^8 + 10C3(0.5)³(0.5)^7

= 1 × (0.5)^10 + 10 × (0.5)^10 + 45 × (0.5)^10 + 120 × (0.5)^10

=0.000977 + 0.00977 + 0.04395 + 0.11719

= 0.171887

3 0

2 years ago

Find all points having an x coordinate of 5 whose distance from the point (2,2) is 5.

stepan [7]

Answer:

(5,6) , (5,-2)

Step-by-step explanation:

Let the required point be P(5,y).

Its distance from (2,2) is 5 units.

So, $(5-2)^{2}$ + $(y-2)^{2}$ = 25.

So, $3^{2}$ + $(y-2)^{2}$ = 25.

9 + $(y-2)^{2}$ = 25.

$(y-2)^{2}$ = 16

Two cases are possible for now

case 1 : y-2 = 4 .

y = 6

required point will be (5,6)

case 2 : y - 2 = -4.

y = -2.

required point will be (5,-2).

Two points are possible : (5,6) , (5,-2).

5 0

2 years ago