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3 years ago

How many solutions can be found for the system of linear equations represented on the graph?

Mathematics

2 answers:

g100num [7]3 years ago

5 0

Answer:

A) no solution

Step-by-step explanation:

These lines are parallel. This means they never intersect; since the solution to a system of equations is the point or points of intersection, this means there is no solution to this system.

skelet666 [1.2K]3 years ago

3 0

Answer: A) no solution

Step-by-step explanation:

The equation of first line = $y=-\dfrac{1}{2}x+4$

which can be written as : $2y=-x+8$

$x+2y=8$

The equation of second line = $y=-\dfrac{1}{2}x-6$

which can be written as : $x+2y=-12$

Since , the ratio of the coefficients of x is equal to the ratio of the coefficients of y but not the ratio of the constants .

$\dfrac{1}{1}=\dfrac{2}{2}\neq\dfrac{8}{-12}$

Therefore , the lines are parallel and do not have any solution.

So , the correct answer is A) no solution.

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Consider a fair coin which when tossed results in either heads (H) or tails (T). If the coin is tossed TWO times 1. List all pos

Volgvan

Answer:

Sample space = {(T,T), (T,H), (HT), (HH)}

Step-by-step explanation:

We are given a fair coin which when tossed one times either gives heads(H) or tails(T).

Now, the same coin is tossed two times.

1) All the possible outcomes

Tails followed by tails

Rails followed by heads

Heads followed by a tail

Heads followed by heads

2) Sample space

{(T,T), (T,H), (HT), (HH)}

3) Formula:

$Probability = \displaystyle\frac{\text{Favourable outcome}}{\text{Total number of outcome}}$

Using the above formula, we can compute the following probabilities.

Probability((T,T)) = $\frac{1}{4}$

Probability((T,H)) = $\frac{1}{4}$

Probability((H,T)) = $\frac{1}{4}$

Probability((H, H)) = $\frac{1}{4}$

Probability(Atleast one tails) = $\frac{3}{4}$

Probability(Atleast one heads) = $\frac{3}{4}$

Probability(Exactly one tails) = $\frac{2}{4}$

Probability(Exactly one heads) = $\frac{2}{4}$

4 0

3 years ago

suppose that 23% of the people have a dog, 18% of the people have a cat and 5% of people own both. what is the probability that

mixer [17]

<h2>Addition</h2><h3>Concept</h3>

Add units together in order to reach a sum.

<h3>Utilization</h3>

In this problem, you are told to find the probability (or %) that someone owns either a dog, OR a cat. If they own both - then they do in fact own a dog or cat. So the formula is:

dog owners % + cat owners % + both owners %

Aka,

$23 + 18 + 5 = 46$

<h3>Answer</h3>

The answer is 46% of people own either a cat or a dog, because 23% + 18% + 5% is 46%.

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3 years ago

Will ran four miles on his first day of training. The next day he ran one-third that distance. How far did he run the second day

igomit [66]

1.3333333333333333333333333333333333333

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3 years ago

Hello, I need help with this, you can put the operation <br><br> (1/2) +7 + (- 2) + (- 3/2) +2

Otrada [13]

-2 and+2 =0
1/2 -3/2= -2/2 or -1
7-1=6

6 0

3 years ago

An engineering crew ran several tests on a new automobile engine they were designing. The mean fuel consumption was 52.4 miles p

zhenek [66]

Any value which is more than 2 standard deviations away from the mean is considered to be "unusual."

2 standard deviations above the mean 52.4 mp would be 52.4+2(1.8), or 56; 2 std devs below the mean would be 52.4 - 2(1.8), or 48.8. Thus, any value larger than 56 or any value smaller than 48.8 would be "unusual."

54.8, 49.1 and 51.3 are not unusual; 56.5 is unusual, because it's greaster than 56.

8 0

3 years ago