Which of these shows line of symmetry?
Select the three correct answers.

Answer: x=3y−9

Step-by-step explanation:Let's solve for x.

−x+3y=9

Step 1: Add -3y to both sides.

−x+3y+−3y=9+−3y

−x=−3y+9

Step 2: Divide both sides by -1.

−x/−1  =  −3y+9/−1

x=3y−9

Answer:

total no of outcomes are '15'

outcomes are shown below.

Step-by-step explanation:

A = {1, 2, 3, 4, 5 ,6 ,7}

subsets which will include {1, 3, 5}

hence, the outcome are :

{1,2,3,5}, {1,3,4,5}, {1,3,5,6}, {1,3,5,7}, {1,2,3,5,6},{1,2,3,5,7},{1,2,3,4,5}, {1,3,4,5,6}, {1,3,4,5,7}, {1,2,3,5,7},{1,2,3,5,6}, {1,3,5,6,7},{1,2,3,4,5,6},{1,2,3,4,5,7}, {1,2,3,4,5,6,7}

so the total no of outcomes are '15'

Answer:

Answer is B

Step-by-step explanation:

If you are unsure about where to start, you could always plot some numbers down until you see a general pattern.

But a more intuitive way is to determine what happens during each transformation.

A regular y = |x| will have its vertex at the origin, because nothing is changed for a y = |x| graph. We have a ray that is reflected at the origin about the y-axis.

Now, let's explore the different transformations for an absolute value graph by taking a y = |x + h| graph.

What happens to the graph?

Well, we have shifted the graph -h units, just like a normal trigonometric, linear, or even parabolic graph. That is, we have shifted the graph h units to its negative side (to the left).

What about the y = |x| + h graph?

Well, like a parabola, we shift it h units upwards, and if h is negative, we shift it h units downwards.

So, if you understand what each transformation does, then you would be able to identify the changes in the shape's location.

Answer:

answer is b

Step-by-step explanation:

because i got it right on E D G E

Answer:

a) The probability that both adults dine out more than once per week = 0.0253

b) The probability that neither adult dines out more than once per week = 0.7069

c) The probability that at least one of the two adults dines out more than once per week = 0.2931

d) Of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.

Step-by-step explanation:

In a sample of 1200 U.S. adults, 191 dine out at a restaurant more than once per week.

Assuming this sample.is a random sample and is representative of the proportion of all U.S. adults, the probability of a randomly picked U.S. adult dining out at a restaurant more than once per week = (191/1200) = 0.1591666667 = 0.1592

Now, assuming this probability per person is independent of each other.

Two adults are picked at random from the entire population of U.S. adults, with no replacement, thereby making sure these two are picked at absolute random.

a) The probability that both adults dine out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult A and adult B dine out more than once per week = P(A n B)

= P(A) × P(B) (since the probability for each person is independent of the other person)

= 0.1592 × 0.1592

= 0.02534464 = 0.0253 to 4 d.p.

b) The probability that neither adult dines out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408

Probability that neither adult dines out more than once per week = P(A' n B')

= P(A') × P(B')

= 0.8408 × 0.8408

= 0.70694464 = 0.7069 to 4 d.p.

c) The probability that at least one of the two adults dines out more than once per week.

Probability that adult A dines out more than once per week = P(A) = 0.1592

Probability that adult A does NOT dine out more than once per week = P(A') = 1 - P(A) = 1 - 0.1592 = 0.8408

Probability that adult B dines out more than once per week = P(B) = 0.1592

Probability that adult B does NOT dine out more than once per week = P(B') = 1 - P(B) = 1 - 0.1592 = 0.8408

The probability that at least one of the two adults dines out more than once per week

= P(A n B') + P(A' n B) + P(A n B)

= [P(A) × P(B')] + [P(A') × P(B)] + [P(A) × P(B)]

= (0.1592 × 0.8408) + (0.8408 × 0.1592) + (0.1592 × 0.1592)

= 0.13385536 + 0.13385536 + 0.02534464

= 0.29305536 = 0.2931 to 4 d.p.

d) Which of the events can be considered unusual? Explain.

The event that can be considered as unusual is the event that has very low probabilities of occurring, probabilities of values less than 5% (0.05).

And of the three events described, the event that can be considered unusual because of its low probability of occurring, 0.0253 (2.53%), is the event that the two randomly selected adults both dine out more than once per week.

Hope this Helps!!!

Answer:

y=-1x-1

Step-by-step explanation:

m=y2-y1/x2-x1

(-3)-1=-4

2-(-2)=4

-4/4=(-1)

y=mx+b

-3=-1(2)+b

-1*2=-2

-2-(-2)=0

(-3)-(-2)=(-1)

b=-1

hope this helps :3

if it did pls mark brainliest