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sdas [7]
3 years ago
15

A point undergoes two transformations. The x-coordinate's sign changes and the y-coordinate increases. What were the transformat

ions?
The point was reflected over the y-axis and translated down.
The point was reflected over the x-axis and translated to the right.
The point was reflected over the y-axis and translated up.
The point was reflected over the x-axis and translated to the left.
Mathematics
2 answers:
aniked [119]3 years ago
6 0
For this case we are going to define the original coordinates:
 (x, y): original coordinates.
 We apply the transformation:
 (x, y) -------> (-x, y + k)
 We have:
 -x: reflection on the y axis.
 y + k: translation k units up.
 Answer:
 
The point was reflected over the y-axis and translated up.
NikAS [45]3 years ago
5 0

Answer:  The point was reflected over the y-axis and translated up.

Step-by-step explanation:

You might be interested in
Pairs that satisfy function y=2x+1
evablogger [386]

Test a pair from each table by substituting their values into the given equation and solving.

A) y = 2x + 1 .......... 2 = 2(0) + 1 .......... 2 ≠ 1

B) y = 2x + 1 .......... 1 = 2(0) + 1 .......... 1 = 1

C) y = 2x + 1 .......... -1 = 2(0) + 1 .......... -1 ≠ 1

D) y = 2x + 1 .......... -2 = 2(0) + 1 .......... -2 ≠ 1


The only pair that satisfied the equation was from answer choice B. Therefore, B is the correct answer.

8 0
3 years ago
A coin is biased such that it results in 2 heads out of every 3 coins flips on average
alina1380 [7]

<span>The mathematical theory of probability assumes that we have a well defined repeatable (in principle) experiment, which has as its outcome a set of well defined, mutually exclusive, events.</span>


If we assume that each individual coin is equally likely to come up heads or tails, then each of the above 16 outcomes to 4 flips is equally likely. Each occurs a fraction one out of 16 times, or each has a probability of 1/16.

Alternatively, we could argue that the 1st coin has probability 1/2 to come up heads or tails, the 2nd coin has probability 1/2 to come up heads or tails, and so on for the 3rd and 4th coins, so that the probability for any one particular sequence of heads and tails is just (1/2)x(1/2)x(1/2)x(1/2)=(1/16).

Now lets ask: what is the probability that in 4 flips, one gets N heads, where N=0, 1, 2, 3, or 4. We can get this just by counting the number of outcomes above which have the desired number of heads, and dividing by the total number of possible outcomes, 16. 
  
 

<span>N     # outcomes with N heads     probability to get N heads</span>

0                1                                       1/16 = 0.0625

1                4                                       4/16 = 1/4 = 0.25

2                6                                      6/16 = 3/8 = 0.375

3                4                                      4/16 = 1/4 = 0.25

4                1                                      1/16 = 0.0625

We can plot these results on a graph as shown below.

 
The dashed line is shown just as a guide to the eye. Notice that the curve has a "bell" shape. The most likely outcome is for N=2 heads, where the curve reaches its maximum value. This is just what you would expect: if each coin is equally likely to land heads as tails, in four flips, half should come up heads, that is N = 4x(1/2) = 2 is the most likely outcome. Note however that an occurrence of N = 1 or N = 3 is not so unlikely - they occur 1/4 or 25% of the time. To have an occurrence of only N = 0, or N = 4 (no heads, or all heads) is much less likely - they occur only 1/16 or 6.25% of the time.

The above procedure is in principle the way to solve all problems in probability. Define the experiment, enumerate all possible mutually exclusive outcomes (which are usually assumed to be each equally likely), and then count the number of these outcomes which have the particular property being tested for (here for example, the number of heads). Dividing this number by the total number of possible outcomes then gives the probability of the system to have that particular property.

Often, however, the number of possible outcomes may be so large that an explicit enumeration would become very tedious. In such cases, one can resort to more subtle thinking to arrive at the desired probabilities. For example, we can deduce the probabilities to get N heads in 4 flips as follows:

N=0: There is only one possible outcome that gives 0 heads, namely when each flip results in a tail. The probability is therefore 1/16.

N=4: There is only one possible outcome that gives 4 heads, namely when each flip results in a head. The probability is therefore 1/16.

N=1: There are 4 possible outcomes which will have only one coin heads. It may be that the 1st coin is heads, and all others are tails; or it may be that the 2nd coin is heads, and all others are tails; or it may be that the 3rd (or the 4th) coin is heads, and all others are tails. Since there are 4 possible outcomes with one head only, the probability is 4/16 = 1/4.

N=3: To get 3 heads, means that one gets only one tail. This tail can be either the 1st coin, the 2nd coin, the 3rd, or the 4th coin. Thus there are only 4 outcomes which have three heads. The probability is 4/16 = 1/4.

N=2: To enumerate directly all the possible outcomes which have exactly 2 heads only, is a bit trickier than the other cases. We will come to it shortly. But we can get the desired probability for N=2 the following way: We have already enumerated all possible outcomes with either N = 0, 1, 3, or 4 heads. These account for 1 + 4 + 4 + 1 = 10 possible outcomes. The only outcomes not include in these 10 are those with exactly N=2 heads. Since there are 16 possible outcomes, and 10 do not have N=2 heads, there must therefore be exactly 16 - 10 = 6 outcomes which do have exactly N=2 heads. The probability for N=2 is therefore 6/16 = 3/8.

2) Consider the experiment of rolling 3 dice, each of which has 6 sides.

What is the probability that no two dice land with the same number side up, i.e. each of the three dice rolls a different number?

Since each die has 6 possible outcomes, the number of possible outcomes for the roll of three dice is 6x6x6 = 216. We could enumerate all these 216 possibilities, and then count the number of outcomes in which each die has a different number. This is clearly too tedious! Instead we reason as follows:


6 0
3 years ago
Read 2 more answers
Mary has used 160 tiles to cover 40% of the back splash in her kitchen how many tiles will it take to cover the entire back spla
dangina [55]

Answer:

400 tiles :)

Step-by-step explanation:

2/5 = 160/x

1/5 = 80/x

80 *5 = 400

6 0
3 years ago
Ive got a big maths test tomorrow ask me questions on fractions or percentages , formula , significant figures , Pythagoras anyt
podryga [215]
What 22/30 times 2/10
7 0
3 years ago
Volume problem. Will give brainliest if correct
svp [43]

Answer:

<em>$ 33.6 to fill this tank, provided a community cost of $2.8 per gallon</em>

Step-by-step explanation:

1. Let us first find the volume of the gas the tank, by the general multiplication of Base * height ⇒ 11 inches * 1.25 feet * 1.75 feet. For the simplicity, we should convert feet ⇒ inches, as such: 1.25 feet = 1.25 * 12 inches = 15 inches, 1.75 feet = 1.75 * 12 inches = 21 inches. Now we have a common unit, let us find the volume ⇒ 11 in. * 15 in. * 21 in. = 3465 inches^3.

2. Let us say that the the average price of gas in my community is $2.8 per gallon. We would first have to convert inches ⇒ gallons provided 1 gallon = 231 inches: 3465/231 = 15 gallons.

4. Now simply multiply this price of 2.8 dollars per gallon by the number of gallons to receive the cost if the tank was full: 2.8 * 15 = <em>$ 42 if this tank was full provided a community cost of $ 2.8 per gallon</em>

5. Now this tank is 20% full, so we must calculate the cost to fill the other 80% up. That would be 80/100 * 42 = 4/5 * 42 = 168/5 = <em>$ 33.6 to fill this tank, provided a community cost of $2.8 per gallon</em>

8 0
3 years ago
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