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3 years ago

Help please it is graphing

Mathematics

1 answer:

Alja [10]3 years ago

4 0

Here are the pairs:

(from the left)

First group: 3rd graph, y = x - 2, table g

Second group: 2nd graph, y = 2x, table b

Third group: 1st graph, y = x + 2, table m

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Solve for the value of s. (35+6)/(25-1)º

o-na [289]

Answer:

615/360º

Step-by-step explanation:

8 0

3 years ago

Complete the table of values for the equation y=x2-2x-1

lord [1]

y = x^2 - 2x - 1

1. If x = -2, then

y = (-2)^2 - 2(-2) - 1

= 4 + 4 - 1

= 7

2. If x = 0, then

y = (0)^2 - 2(0) - 1

= 0 - 0 - 1

= -1

3. If x = 2, then

y = (2)^2 - 2(2) - 1

= 4 - 4 - 1

= -1

4. If x = 3, then

y = (3)^2 - 2(3) - 1

= 9 - 6 - 1

= 2

Hope you could understand.

If you have any query, feel free to ask.

8 0

3 years ago

If I am in Vermont, then I am in the North.

raketka [301]

The correct answer is D.

This is because a converse statement switches the places of the hypothesis and conclusion.

6 0

3 years ago

The weights of steers in a herd are distributed normally. The standard deviation is 300lbs and the mean steer weight is 1100lbs.

gizmo_the_mogwai [7]

Answer:

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(920≤ x≤1730) = 0.7078

Step-by-step explanation:

Step(i):-

Given mean of the Population = 1100 lbs

Standard deviation of the Population = 300 lbs

Let 'X' be the random variable in Normal distribution

Let x₁ = 920

$Z = \frac{x-mean}{S.D} = \frac{920-1100}{300} = - 0.6$

Let x₂ = 1730

$Z = \frac{x-mean}{S.D} = \frac{1730-1100}{300} = 2.1$

Step(ii)

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(x₁≤ x≤x₂) = P(Z₁≤ Z≤ Z₂)

= P(-0.6 ≤Z≤2.1)

= P(Z≤2.1) - P(Z≤-0.6)

= 0.5 + A(2.1) - (0.5 - A(-0.6)

= A(2.1) +A(0.6) (∵A(-0.6) = A(0.6)

= 0.4821 + 0.2257

= 0.7078

Conclusion:-

The probability that the weight of a randomly selected steer is between 920 and 1730 lbs

P(920≤ x≤1730) = 0.7078

5 0

3 years ago

Are two lines parallel, perpendicular or neither 3x+7y=15 and 7x-3y=6

maw [93]

Answer:

The lines are perpendicular

Step-by-step explanation:

we know that

If two lines are parallel, then their slopes are equal

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

we have

$3x+7y=15$ ----> equation A

isolate the variable y

$7y=-3x+15$

Divide by 7 both sides

$y=-\frac{3}{7}x+\frac{15}{7}$

The slope of the line A is $m_A=-\frac{3}{7}$

$7x-3y=6$ ----> equation B

isolate the variable y

$3y=7x-6$

Divide by 3 both sides

$y=\frac{7}{3}x-2$

The slope of the line B is $m_B=\frac{7}{3}$

Compare the slope of both lines

$m_A=-\frac{3}{7}$

$m_B=\frac{7}{3}$

$m_A \neq m_B$ ----> the lines are not parallel

Find the product of the slopes

$-\frac{3}{7}(\frac{7}{3})=-1$ ----> the lines are perpendicular

4 0

4 years ago