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bezimeni [28]
2 years ago
11

Marissa wants to write an abbreviated set of directions for finding coordinates of a figure reflected across the y-axis. Which m

apping notation is correct?
1. (x,y) --> (-x,y)
2. (x,y) --> (x,-y) 
3. (x,y) --> (-x,-y) 
4. (x,y) --> (y,x) 
Mathematics
2 answers:
Sedaia [141]2 years ago
7 0
The answer is the very first one! 
(x,y) --> (-x,y) 

It demonstrates a perfect reflection across the y-axis. 

Good Luck! 
Nutka1998 [239]2 years ago
7 0

Answer:

1. (x,y) --> (-x,y)

Step-by-step explanation:

We are given that,

Marissa needs to write the rule for finding the co-ordinates of a figure reflected across the y-axis.

We know,

Reflection flips the figure over the line of reflection.

So, the reflection of a figure over y-axis will flip the figure over y-axis.

<em>Thus, we get that there will be no change in the y-co-ordinate and the x co-ordinate will be multiplied by a negative sign.</em>

So, the co-ordinates can be found out by the rule,

(x,y)   ----------- >   (-x,y).

Hence, option 1 is correct.

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Here are Babe Ruth's home run totals for the New York Yankees during the years 1920 - 1934:54, 59, 35, 41, 46, 25, 47, 60, 54, 4
Irina-Kira [14]

Given data is 54, 59, 35, 41, 46, 25, 47, 60, 54, 46, 49, 46, 41, 34, 22

Arranging it in increasing order:

22 25 34 35 41 41 46 46 46 47 49 54 54 59 60

Number of data items, n=15.

Finding Median, First Quartile, and Third Quartile

Finding\;\;median\\M=\frac{n+1}{2}th\;\;term =\frac{15+1}{2}=8th\;\;term\;\; i.e. \;\;46

22 25 34 35 41 41 46 46 46 47 49 54 54 59 60

Now finding First Quartile from set: 22 25 34 35 41 41 46

Q₁ = middle term i.e. 35

Now finding Third Quartile from set: 46 47 49 54 54 59 60

Q₃ = middle term i.e. 54

So we have Q₁ = 35, M = 46, and Q₃ = 54. Lower bound is 22 and Upper bound is 60.

While drawing the given data on number line and marking First quartile, Median, Third quartile on number line. We can clearly observe that option A matches with given data points.

Hence, option A i.e. Box-plot A is the correct choice.

5 0
2 years ago
Read 2 more answers
The weekly salary paid to employees of a small company that supplies​ part-time laborers averages ​$750 with a standard deviatio
poizon [28]

Answer:

(a) The fraction of employees is 0.84.

(b)

\mu=850\\\\\sigma=450

(c)

\mu=787.5\\\\\sigma=472.5

(d) No. The left part of the distribution would be truncated too much.

Step-by-step explanation:

(a) If the weekly salaries are normally​ distributed, estimate the fraction of employees that make more than ​$300 per week.

We have to calculate the z-value and compute the probability

z=\frac{X-\mu}{\sigma}= \frac{300-750}{450}=\frac{-450}{450}=-1\\\\P(X>300)=P(z>-1)=0.84

(b) If every employee receives a​ year-end bonus that adds ​$100 to the paycheck in the final​ week, how does this change the normal model for that​ week?

The mean of the salaries grows $100.

\mu_{new}=E(x+C)=E(x)+E(C)=\mu+C=750+100=850

The standard deviation stays the same ($450)

\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(x+C)-(\mu+C)]^2}  } =\sqrt{\frac{1}{N} \sum{(x+C-\mu-C)^2}  }\\\\ \sigma_{new}=\sqrt{\frac{1}{N} \sum{(x-\mu)^2}  } =\sigma

(c) If every employee receives a 5​% salary increase for the next​ year, how does the normal model​ change?

The increases means a salary X is multiplied by 1.05 (1.05X)

The mean of the salaries grows 5%, to $787.5.

\mu_{new}=E(ax)=a*E(x)=a*\mu=1.05*750=487.5

The standard deviation increases by a 5% ($472.5)

\sigma_{new}=\sqrt{\frac{1}{N} \sum{[(ax)-(a\mu)]^2}  } =\sqrt{\frac{1}{N} \sum{a^2(x-\mu)^2}  }\\\\ \sigma_{new}=\sqrt{a^2}\sqrt{\frac{1}{N} \sum{(x-\mu)^2}}=a*\sigma=1.05*450=472.5

(d) If the lowest salary is ​$300 and the median salary is ​$525​, does a normal model appear​ appropriate?

No. The left part of the distribution would be truncated too much.

7 0
1 year ago
The following data represent exam scores in a statistics class taught using traditional lecture and a class taught using a "flip
lapo4ka [179]

The table of this question is in the attachments.

Answer and Step-by-step explanation: <u>Standard</u> <u>deviation</u> is a measure of how spread the data is from the mean. It is calculated as:

s = √∑(x - μ)² / n - 1

where μ is the mean of the set.

<u>Range</u> is the difference between the highest and lowest value of a data set.

(a) <u>Range of Traditional course</u>:

range = 80.4 - 56

range = 24.4

<u>Range of "flipped" course</u>:

range = 92.6 - 63.5

range = 29.1

Comparing ranges, the "flipped" course has more dispersion than the traditional.

(b) <u>Standard Deviation of Traditional course</u>:

mean = 71.6

s = \sqrt{\frac{(70.3 - 71.6)^{2}+...+(59.1-71.6)^{2}}{13-1}

s = 8.95

<u>Standard Deviation of "flipped" course</u>:

mean = 77.6

s = \sqrt{\frac{(75.5 - 77.6)^{2}+...+(76.9-77.6)^{2}}{13-1}

s = 8.3

Comparing standard deviation, traditional course has more dispersion.

(c) If you change one score, range for traditional will be:

range = 591 - 56

range = 535

Changing one score increase in almost 22 times the range for this category.

7 0
1 year ago
The center of a hyperbola is located at (0, 0). One focus is located at (0, 5) and its associated directrix is represented by th
Lady bird [3.3K]
For a hyperbola   \dfrac{y^{2}}{a^{2}}-\dfrac{x^{2}}{b^{2}}=1
where   a^{2}+b^{2}=c^{2}
the directrix is the line   y=\dfrac{a^{2}}{c}
and the focus is at (0, c).

Here, we have c = 5, a² = 9, so b² = 5² - 9 = 16.
  a = √9 = 3
  b = √16 = 4

Your hyperbola's constants are ...
  a = 3
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______
Please note that the equation of a hyperbola has a negative sign for one of the terms. The equation given in your problem statement is that of an ellipse.
8 0
2 years ago
Read 2 more answers
In △ABC, m∠ABC=40°,
Alchen [17]

Let m∠CLN = x. Then m∠ALM = 3x, and m∠A = 90°-x, m∠C = 90°-3x.

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... 180° = 40° + (90° -x) + (90° -3x)

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The ratio of CN to CL is

... CN/CL = cos(∠C) = cos(60°)

... CN/CL = 1/2

so ...

... CN = (1/2)CL

5 0
2 years ago
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