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AlexFokin [52]
3 years ago
9

Given a triangle JLM with vertices J(-4,3), L(-2,7), and M(2,7). What would the coordinates be if it was reflected across the li

ne y=1?
Mathematics
1 answer:
topjm [15]3 years ago
3 0

Answer: M'(2, - 5), L'(-2, -5), j'(-4, - 1)

Step-by-step explanation:

When we do a reflection over a given line, the distance between all the points (measured perpendicularly to the line) does not change.

The line is y = 1.

Notice that a reflection over a line y = a (for any real value a) only changes the value of the variable y.

Let's reflect the points:

J(-4, 3)

The distance between 3 and 1 is:

D = 3 - 1 = 2.

Then the new value of y must also be at a distance 2 of the line y = 1

1 - 2 = 1

The new point is:

j'(-4, - 1)

L(-2, 7)

The distance between 7 and 1 is:

7 - 1 = 6.

The new value of y will be:

1 - 6 = -5

The new point is:

L'(-2, -5)

M(2,7)

Same as above, the new point will be:

M'(2, - 5)

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6 people sharing 40 lbs of gold evenly. How many do they get? Then subtract 1 2/3 from the result of the problem above & sim
r-ruslan [8.4K]

Answer: 6

Step-by-step explanation:

Given Data:

Number of people = 6

Amount of Gold to be shared = 40lbs

Subtract 2/3 from your answer.

Therefore

If six people are to share 40lbs of gold evenly

= 40lbs / 6

Divide through by 2

= 20lbs / 3

= 20lbs / 3 - 2/3

= 18/3

= 6

Each person would get approximately 6lbs of gold

7 0
3 years ago
Read 2 more answers
Are the marks one receives in a course related to the amount of time spent studying the subject? To analyze this mysterious poss
sertanlavr [38]

Answer:

a) 98.522

b) 0.881

c) The correlation coefficient and co-variance shows that there is positive association between marks and study time. The correlation coefficient suggest that there is strong positive association between marks and study time.

Step-by-step explanation:

a.

As the mentioned in the given instruction the co-variance is first computed in excel by using only add/Sum, subtract, multiply, divide functions.

Marks y Time spent x y-ybar x-xbar (y-ybar)(x-xbar)

77                    40 5.1         1.3 6.63

63                     42 -8.9            3.3 -29.37

79                     37 7.1            -1.7 -12.07

86                     47 14.1            8.3 117.03

51                    25 -20.9  -13.7 286.33

78                     44 6.1            5.3 32.33

83                      41 11.1            2.3 25.53

90                     48 18.1            9.3 168.33

65                     35 -6.9           -3.7 25.53

47                    28 -24.9 -10.7 266.43

Covariance=\frac{sum[(y-ybar)(x-xbar)]}{n-1}

Co-variance=886.7/(10-1)

Co-variance=886.7/9

Co-variance=98.5222

The co-variance computed using excel function COVARIANCE.S(B1:B11,A1:A11) where B1:B11 contains Time x column and A1:A11 contains Marks y column. The resulted co-variance is 98.52222.

b)

The correlation coefficient is computed as

Correlation coefficient=r=\frac{sum[(y-ybar)(x-xbar)]}{\sqrt{sum[(x-xbar)]^2sum[(y-ybar)]^2} }

(y-ybar)^2 (x-xbar)^2

26.01        1.69

79.21       10.89

50.41             2.89

198.81       68.89

436.81       187.69

37.21       28.09

123.21        5.29

327.61        86.49

47.61         13.69

620.01         114.49

sum(y-ybar)^2=1946.9

sum(x-xbar)^2=520.1

Correlation coefficient=r=\frac{886.7}{\sqrt{520.1(1946.9)} }

Correlation coefficient=r=\frac{886.7}{\sqrt{1012582.69} }

Correlation coefficient=r=\frac{886.7}{1006.2717 }

Correlation coefficient=r=0.881

The correlation coefficient computed using excel function CORREL(A1:A11,B1:B11) where B1:B11 contains Time x column and A1:A11 contains Marks y column. The resulted correlation coefficient is 0.881.

c)

The correlation coefficient and co-variance shows that there is positive association between marks and study time. The correlation coefficient suggest that there is strong positive association between marks and study time. It means that as the study time increases the marks of student also increases and if the study time decreases the marks of student also decreases.

The excel file is attached on which all the related work is done.

Download xlsx
7 0
3 years ago
Johnny invested $1000 in savings account that yields 2% interest annually. How much will Johnny's investment be worth in 10 year
Ann [662]
Calculate the interest in a year
Multiply the annual percentage of interest by the investment to find the interest.
interest in a year = 2% × 1,000
interest in a year = 0.02 × 1,000
interest in a year = 20
The annual interest is $20

Calculate the interest after 10 years
Multiply the annual interest by 10
interest 10 years = 20 × 10
interest 10 years = 200
The sum of interest after 10 years is $200

Find total investment
investment = first investment + interest in 10 years
investment = 1,000 + 200
investment = 1,200

The investment will be worth $1,200 in 10 years
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3. Suppose your bank honors a check for which you don't have sufficient funds in your checking account. This action means that y
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