find the slope of the line on the graph write your answer as a fraction or a whole number not a mixed number or decimal

Which system of equations is inconsistent?

Agata [3.3K]

Answer:

Option A is correct.

The system of equation is inconsistent is;

2x+8y=6

5x+20y=2

Explanation:

* A system of equations is called an inconsistent system, if there is no solution because the lines are parallel.

* If a system has at least one solution, it is said to be consistent .

*A dependent system of equations is when the same line is written in two different forms so that there are infinite solutions.

(A)

2x+8y=6

5x+20y=2

This is inconsistent, because as shown below in the graph of figure 1 that the lines do not intersect, so the graphs are parallel and there is no solution.

(B)

5x+4y=-14

3x+6y=6

this system of equation is Consistent because it has exactly one solution as shown below in the graph of Figure 2 and also it is independent.

(C)

x+2y=3

4x+6y=5

this system of equation is Consistent because it has exactly one solution as shown below in the graph of Figure 3.

(D)

3x-2y=2

6x-4y=4

this is a consistent system and has an infinite number of solutions, it is dependent because both equations represent the same line. as shown below in the graph of Figure-4.

Therefore, the only Option A system of equation is inconsistent.

8 0

2 years ago

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The Parker family drove a total of 180 miles on the road trip. This distance is 1.5 times the distance they drove on the first d

Veronika [31]

Parker family drove 120 miles on the first day

Step-by-step explanation:

Given:

Total miles driven on the road trip= 180 miles

solution:

Lets assume the distance drove on the first day is x

The total distance covered on the road trip is 180 miles. so,

$(x) *1.5 = 180$

$x=\frac{180}{1.5}$

Hence, x = 120.

The distance travelled on the first day is 120 miles.

4 0

2 years ago

A random sample of 50 units is drawn from a production process every half hour. the true fraction of nonconforming products manu

Naya [18.7K]

solution:

The probability mass function for binomial distribution is,

Where,

X=0,1,2,3,…..; q=1-p

find the probability that (p∧ ≤ 0.06) , substitute the values of sample units (n) , and the probability of nonconformities (p) in the probability mass function of binomial distribution.

Consider x to be the number of non-conformities. It follows a binomial distribution with n being 50 and p being 0.03. That is,

binomial (50,0.02)

Also, the estimate of the true probability is,

p∧ = x/50

The probability mass function for binomial distribution is,

Where,

X=0,1,2,3,…..; q=1-p

The calculation is obtained as

P(p^ ≤ 0.06) = p(x/20 ≤ 0.06)

= 50cx ₓ (0.03)x ₓ (1-0.03)50-x

= (50c0 ₓ (0.03)0 ₓ (1-0.03)50-0 + 50c1(0.03)1 ₓ (1-0.03)50-1 + 50c2 ₓ (0.03)2 ₓ (1-0.03)50-2 +50c3 ₓ (0.03)3 ₓ (1- 0.03)50-3 )

=( ₓ (0.03)0 ₓ (1-0.03)50-0 + ₓ (0.03)1 ₓ (1-0.03)50-1 + ₓ (0.03)2 ₓ (1-0.03)50-2 ₓ (0.03)3 ₓ (1-0.03)50-3 )

5 0

2 years ago

Find the measure of angle x in the figure below: (1 point)

nignag [31]

50 degrees is the answer

4 0

2 years ago

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In high-school 135 freshmen were interviewed.

timama [110]

Answer:

a) n(none) = 25

b) n(PE but not Bio) = 25

c) n(ENG but not both BIO and PE) = 55

d) n(students that did not take Eng or Bio) = 40

e) P( Students did not take exactly two subjects) = 0.65

Step-by-step explanation:

From the Venn diagram drawn:

a) Number of students that took none

n(Freshmen) = 135

n(all three) = 5

n (PE and Bio) = 10

n(PE and Eng) = 15

n(Bio and Eng) = 7

n (PE and Bio only) = 10 - 5 = 5

n(PE and Eng only) = 15 - 5 = 10

n(Bio and Eng only) = 7 - 5 = 2

n(PE only) = 35 - 5 - 5 - 10 = 15

n(Bio only) = 42 - 5 - 5 - 2 = 30

n(Eng only) = 60 - 10 - 5 -2 = 43

n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)

135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)

135 = 110 + n(none)

n(none) = 135 - 110

n(none) = 25

b)Number of students that too PE but not Bio

n(PE but not bio)= n(PE only) + n(PE and Eng only)

n(PE but not Bio) = 15 + 10

n(PE but not Bio) = 25

c) Number of students that took ENG but not both BIO and PE

n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10

n(ENG but not both BIO and PE) = 55

d) Number of students that did not take ENG or BIO

n( students that did not take Eng or Bio) = n(PE only) + n(none)

n(students that did not take Eng or Bio) = 15 + 25

n(students that did not take Eng or Bio) = 40

e) Probability that a randomly-chosen student from this group did not take exactly two subjects

n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)

n( Students that did not take exactly two subjects) = 15 + 30 + 43

n( Students that did not take exactly two subjects) = 88

P( Students did not take exactly two subjects) = 88/135

P( Students did not take exactly two subjects) = 0.65

3 0

2 years ago

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