LUUK al uit grapii velow.
Part B
-4
Part A
-3-2
Part C
3
2
-2
-3
Part D
Which part of the graph best represents the solution set to the system of
inequalities y ≥x+1 and y + x>-1? (5 points)

LUUK Al Uit Grapii Velow.Part B-4Part A-3-2Part C32-2-3Part DWhich Part Of The Graph Best Represents

Answers

Answer 1

The solution set of given inequalities are represented by Part A.

The given inequalities are

⇒ y ≥ x + 1 and y + x > -1

Hence, The related equations of both inequalities are

y = x + 1

Put x=0, to find the y-intercept and put y=0, to find x intercept.

y = 0 + 1

y = 1

And, 0 = x + 1

x = - 1

Therefore, x-intercept of the equation is (-1,0) and y-intercept is (0,1).

Similarly, for the second related equation

y + x = - 1

y + 0 = - 1

y = - 1

0 + x = - 1

x = - 1

Therefore x-intercept of the equation is (-1,0) and y-intercept is (0,-1).

Now, join the x and y-intercepts of both lines to draw the line.

Now check the given inequalities by (0,0).

0 ≥ 0 + 1

0 ≥ 1

It is a false statement, therefore the shaded region is in the opposite side of origin.

0 + 0 ≥ - 1

0 ≥ - 1

It is a true statement, therefore the shaded region is about the origin.

Hence, From the below figure we can say that the solution set of given inequalities are represented by Part A.

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LUUK Al Uit Grapii Velow.Part B-4Part A-3-2Part C32-2-3Part DWhich Part Of The Graph Best Represents

Related Questions

Teena uses 1/4 cup of oil for a cake. How many cakes can she make if she has 6 cups of oil?

Answers

Answer:

24 cakes.

Step-by-step explanation:

6 cups of oil divided by 1/4 cup oil per cake = 24 cakes

6/(1/4) = 24

or 6/(0.25) = 24

She can make 24 cakes with 6 cups of oil.

Instructions: Find the missing probability.

P(B)=1/2P(A|B)=11/25P(AandB)=

Answers

We can use the formula:

P(A and B) = P(B) x P(A|B)

We are given:

P(B) = 1/2

P(A|B) = 11/25

Substituting these values into the formula, we get:

P(A and B) = (1/2) x (11/25) = 11/50

Therefore, P(A and B) = 11/50.

Please help. Is the answer even there?

Answers

The critical values t₀ for a two-sample t-test is ± 2.0.6

To find the critical values t₀ for a two-sample t-test to test the claim that the population means are equal (i.e., µ₁ = µ₂), we need to use the following formula:

t₀ = ± t_(α/2, df)

where t_(α/2, df) is the critical t-value with α/2 area in the right tail and df degrees of freedom.

The degrees of freedom are calculated as:

df = (s₁²/n₁ + s₂²/n₂)² / [(s₁²/n₁)²/(n₁-1) + (s₂²/n₂)²/(n₂-1)]

n₁ = 14, n₂ = 12, X₁ = 6,X₂ = 7, s₁ = 2.5 and s₂ = 2.8

α = 0.05 (two-tailed)

First, we need to calculate the degrees of freedom:

df = (s₁²/n₁ + s₂²/n₂)² / [(s₁²/n₁)²/(n₁-1) + (s₂²/n₂)²/(n₂-1)]

= (2.5²/14 + 2.8²/12)² / [(2.5²/14)²/13 + (2.8²/12)²/11]

= 24.27

Since this is a two-tailed test with α = 0.05, we need to find the t-value with an area of 0.025 in each tail and df = 24.27.

From a t-distribution table, we find:

t_(0.025, 24.27) = 2.0639 (rounded to four decimal places)

Finally, we can calculate the critical values t₀:

t₀ = ± t_(α/2, df) = ± 2.0639

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