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I am Lyosha [343]
2 years ago
10

Use the elimination method to solve each linear system.

Mathematics
1 answer:
Free_Kalibri [48]2 years ago
7 0

Answer:

1) x=4, y=7

2) x=1, y=2

3) x=-3, y=2

4) x=1, y=1

word problem) 5g of nutmeg and 20g of cinnamon

Step-by-step explanation:

1)

2x-9y=-55

2(x+2y) = 2x+4y = 36

(2x+4y)-(2x-9y)=36-(-55)

13y = 91

y = 7 so x = 4

2)

2(3x+2y) = 6x+4y = 14

6x-6y=-6

(6x+4y)-(6x-6y) = 14-(-6)

10y = 20

y = 2 so x = 1

3)

2(6x+10y) = 12x+20y = 4

5(7x+4y) = 35x+20y = -65

(35x+20y) - (12x+20y) = -65-4

23x = -69

x = -3 so y = 2

4)

just add them

(3x+2y)+(x-2y) = 5+(-1)

4x = 4

x = 1 so y= 1

last)

c+n = 25

9c + 12.5n = 25*9.7 = 242.5

9(c+n) = 9*25

9c+9n=225

9c+12.5n - 9c-9n = 242.5-225

3.5n = 17.5

n = 5 so c = 20

5g of nutmeg and 20g of cinnamon

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Answer:

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

Step-by-step explanation:

For each bridge, there are only two possible outcomes. Either it has rating of 4 or below, or it does not. The probability of a bridge being rated 4 or below is independent from other bridges. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

For the year 2020, the engineers forecast that 9% of all major Denver bridges will have ratings of 4 or below.

This means that p = 0.09

Use the forecast to find the probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

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So

P(X < 4) + P(X \geq 4) = 1

We want P(X \geq 4)

So

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{12,0}.(0.09)^{0}.(0.91)^{12} = 0.3225

P(X = 1) = C_{12,1}.(0.09)^{1}.(0.91)^{11} = 0.3827

P(X = 2) = C_{12,2}.(0.09)^{2}.(0.91)^{10} = 0.2082

P(X = 3) = C_{12,3}.(0.09)^{3}.(0.91)^{9} = 0.0686

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.3225 + 0.3827 + 0.2082 + 0.0686 = 0.982

Finally

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.982 = 0.0180

1.80% probability that in a random sample of 12 major Denver bridges, at least 4 will have an inspection rating of 4 or below in 2020.

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