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ololo11 [35]
3 years ago
7

The sum of Sally and Beth's age is 60. Six years ago, Sally was 3 times as old as Beth. What are their present ages?

Mathematics
2 answers:
Thepotemich [5.8K]3 years ago
4 0
Sally is 45 and Beth is 15.

45 + 15 = 60

15*3 = 45
pav-90 [236]3 years ago
3 0

Answer: Present age of Beth is 24 years.

And Present age of Sally becomes 36 years.

Step-by-step explanation:

Let the present age of Beth be 'x'.

Six years ago,

Age of Beth becomes (x-6) years

Age of Sally becomes

3(x-6)\ yrs

Sum of Sally and Beth's age = 60

According to question, it becomes

x-6+3(x-6)=60+6+6\\\\x-6+3x-18=72\\\\4x-24=72\\\\4x=72+24\\\\4x=96\\\\x=\frac{96}{4}=24\ yrs

So, Present age of Beth is 24 years.

And Present age of Sally becomes 60-24=36 years.

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Answer:

m=\frac{S_{xy}}{S_{xx}}  

Where:  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}  

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}  

\bar x= \frac{\sum x_i}{n}  

\bar y= \frac{\sum y_i}{n}  

And we can find the intercept using this:  

b=\bar y -m \bar x  

On this case the correct answer would be:

E.  none of the above

Since the intercept has no association between the increase/decrease of the dependent variable respect to the independent variable

Step-by-step explanation:

Assuming the following options:

A.  there is a positive correlation between X and Y

B.  there is a negative correlation between X and Y

C.  if X is increased, Y must also increase

D.  if Y is increased, X must also increase

E.  none of the above

If we want a model y = mx +b where m represent the lope and b the intercept

m=\frac{S_{xy}}{S_{xx}}  

Where:  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}  

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}  

\bar x= \frac{\sum x_i}{n}  

\bar y= \frac{\sum y_i}{n}  

And we can find the intercept using this:  

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On this case the correct answer would be:

E.  none of the above

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3 years ago
Which is the graph of y =(x) - 2?
Likurg_2 [28]

Answer: i dont see the graphs

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Item 7
Mariulka [41]

Answer:

A = 74.7^\circ

B = 42.5^\circ

C = 62.8^\circ

Step-by-step explanation:

Given

A = (-1,2) \to (x_1,y_1)

B = (2,8) \to (x_2,y_2)

C = (4,1) \to (x_3,y_3)

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2

For AB

A = (-1,2) \to (x_1,y_1)

B = (2,8) \to (x_2,y_2)

d = \sqrt{(-1 - 2)^2 + (2 - 8)^2

d = \sqrt{(-3)^2 + (-6)^2

d = \sqrt{45

So:

AB = \sqrt{45

For BC

B = (2,8) \to (x_2,y_2)

C = (4,1) \to (x_3,y_3)

BC = \sqrt{(2 - 4)^2 + (8 - 1)^2

BC = \sqrt{(-2)^2 + (7)^2

BC = \sqrt{53

For AC

A = (-1,2) \to (x_1,y_1)

C = (4,1) \to (x_3,y_3)

AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2

AC = \sqrt{(-5)^2 + (1)^2

AC = \sqrt{26

So, we have:

AB = \sqrt{45

BC = \sqrt{53

AC = \sqrt{26

By representation

AB \to c

BC \to a

AC \to b

So, we have:

a = \sqrt{53

b = \sqrt{26

c = \sqrt{45

By cosine laws, the angles are calculated using:

a^2 = b^2 + c^2 -2bc \cos A

b^2 = a^2 + c^2 -2ac \cos B

c^2 = a^2 + b^2 -2ab\ cos C

a^2 = b^2 + c^2 -2bc \cos A

(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A

53 = 26 +45 - 2 * 34.21 * \cos A

53 = 26 +45 - 68.42 * \cos A

Collect like terms

53 - 26 -45 = - 68.42 * \cos A

-18 = - 68.42 * \cos A

Solve for \cos A

\cos A =\frac{-18}{-68.42}

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Take arc cos of both sides

A =\cos^{-1}(0.2631)

A = 74.7^\circ

b^2 = a^2 + c^2 -2ac \cos B

(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B

26 = 53 +45 -97.67 * \cos B

Collect like terms

26 - 53 -45= -97.67 * \cos B

-72= -97.67 * \cos B

Solve for \cos B

\cos B = \frac{-72}{-97.67}

\cos B = 0.7372

Take arc cos of both sides

B = \cos^{-1}(0.7372)

B = 42.5^\circ

For the third angle, we use:

A + B + C = 180 --- angles in a triangle

Make C the subject

C = 180 - A -B

C = 180 - 74.7 -42.5

C = 62.8^\circ

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Colt1911 [192]

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Attached is a screenshot.

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