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2 years ago

Which transformations can be used to carry ABCD onto itself? Check all that applyPLEASE HELP

Mathematics

2 answers:

Tju [1.3M]2 years ago

6 0

Answer:

Options A and B.

Step-by-step explanation:

Rectangle ABCD will overlap itself when reflected by a straight line y = 2, which passes through the middle of the side AD.

After reflection point A will overlap the point D.

Therefore, option A will be the correct option.

Similarly when this rectangle is rotated clockwise or counterclockwise by the center of ABCD at 180°, then the point point B will overlap A.

And the shape of rectangle will remain the same.

Therefore, Option B is also correct.

eduard2 years ago

3 0

Answer:

A and B

Step-by-step explanation:

Reflection and rotation

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A hockey coach recorded the number of shots taken by the home team and the number taken by the visiting team in 20 games. He dis

Harrizon [31]

Answer:

The most accurate inference from this boxplot is that: The visiting team had more variability in the number of shots taken.

Step-by-step explanation:

The box and whiskers plot is a way of presenting data that gives 5 major information about the distribution

From the whiskers of the plot, one can read

- The minimum value

- The maximum value

Then, from the boxplot, one can read

- The Median, represented by the middle line of the boxplot.

- The first Quartile or 25th percentile, represented by the lower end of the boxplot.

- The third quartile or 75th percentile, represented by the upper end of the boxplot.

Other variables that can be obtained from these five data points include

- The range of the distribution (maximum value minus minimum value)

- The interquartile range (a measure of variation, which is the difference between the third and first quartile of the distribution)

For the two boxplots that the coach made

Home team

Whiskers range from 16 to 32

Minimum value = 16

maximum value = 32

the box ranges from 20 to 23. A line divides the box at 22.

First quartile = 20

Third quartile = 23

Median = 22

IQR = 23 - 20 = 3

For the visiting team

Whiskers range from 14 to 32

Minimum value = 14

maximum value = 32

the box ranges from 16 to 24. A line divides the box at 18.

First quartile = 16

Third quartile = 24

Median = 18

IQR = 24 - 16 = 8

Since the median only represents the midpoint of the distribution, one cannot conclude with certainty that home team took more shots than the visiting team, information on the mean would confirm that.

A less controversial and evident inference is that the visiting team had more variability in their shots as their distribution has a higher Interquartile Range (IQR of 8 > 3) which is a direct measure of variation for distributions.

Hope this Helps!!!

7 0

2 years ago

Read 2 more answers

1.nadia selects three marbles from the jar to the right without without replacing them.what is the probability that all three ma

rjkz [21]

32marbles in the jar .

5 0

2 years ago

Read 2 more answers

Norma Jean makes $25 per hour. She works 35 hours per week. She gets a commission of 15% on her total sales. How much should Nor

TEA [102]

Answer:

$24166.67

Step-by-step explanation:

$25 per hour and works 35 hours per week

Norma Jean makes 35 * 25 (without sales) = $875

Total commission made by Norma = 4500-875 = $3625

let total number of sales by Norma be x

15% is the same as 15/100

so (15/100) * x =3625 multiplying both sides by 100

15x = 362500 making x subject of formula

x = 362500/15

x = 24 166.67

Norma should sell total items worth $24166.67

6 0

2 years ago

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At a gymnastics meet, three judges evaluate the balance beam performances of five gymnasts, using a scale from 1 to 10, where 10

Art [367]

Answer:

As 2.551 < 3.84 therefore we reject H0.

As 44.803 > 4.46 so we accept null hypothesis.

Step-by-step explanation:

The answer is attached.

There are three judges so v1 = 3-1= 2 and v2 = (4*2)= 8

There are five gymnasts so v1 = 5-1= 4 and v2= (4*2)= 8

For alpha = 0.05 we find the value of F1 and F2 from the table.

Download docx

6 0

2 years ago

An attempt to establish a video call via some social media app may fail with probability 0.1. If connection is established and i

xxMikexx [17]

Answer:

(1). y = x ~ Exp (1/3).

(2). Check attachment.

(3). EY = 3(1 - e^-2).

(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.

Step-by-step explanation:

Kindly check the attachment to aid in understanding the solution to the question.

So, from the question, we given the following parameters or information or data;

(A). The probability in which attempt to establish a video call via some social media app may fail with = 0.1.

(B). " If connection is established and if no connection failure occurs thereafter, then the duration of a typical video call in minutes is an exponential random variable X with E[X] = 3. "

(C). "due to an unfortunate bug in the app all calls are disconnected after 6 minutes. Let random variable Y denote the overall call duration (i.e., Y = 0 in case of failure to connect, Y = 6 when a call gets disconnected due to the bug, and Y = X otherwise.)."

(1). Hence, for FY(y) = y = x ~ Exp (1/3) for the condition that zero is equal to y = x < 6.

(2). Check attachment.

(3). EY = 3(1 - e^-2).

(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.

The condition to follow in order to solve this question is that y = 0 if x ≤ 0, y = x if 0 ≤ x ≤ 6 and y = 6 if x ≥ 6.

8 0

2 years ago