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Xelga [282]
3 years ago
15

Find three consecutive integers whose sum is 57????

Mathematics
1 answer:
Setler79 [48]3 years ago
7 0

Answer:18 19 20

Step-by-step explanation:"Consecutive"  means that the integers will follow each other in value, for example:  1, 2, 3 or 4, 5, 6.  Also, no decimals are needed here because "integers" are whole, counting numbers. Here is the set up:   Let x= the first integer     Then    X+1= 2nd consecutive integer   and x+2= 3rd  .  

Suppose that x=1   x+1= 1+1=2   and x+2=1+2=3   However, you need specific consecutive numbers whose sum is 57.  Remember that sum means to add:

x+  (x+1)  + (x+2) = 57                 Addition of all 3 consecutive numbers   Now solve for x

                                                   and substitute into each part to come up with the three integers:

3x + 3= 57        3x=54               x=54/3  =  18            x=18,   x+1= 18+1=19      x+2=18+2=20

Check your answer:  18+19+20=57                 57=57  Check

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In right triangle &lt;ABC,<br>M&lt;A = 42°. Find m&lt;B.​
wlad13 [49]

Answer:

m<B is 48°

Step-by-step explanation:

m<C is already 90°

m<A is 42°

42+90+x=180

132+x=180

x=48°

5 0
3 years ago
Given the following data: X 1 4 6 7 Y 9 7 8 1 a) Find the coefficient of correlation. b) Find the equation of the regression lin
Viktor [21]

Answer:

a) r=-0.719

b) y=10.642-0.976x

c)Predicted y=7.714

Step-by-step explanation:

a)

sumx=1+4+6+7=18

sumy=9+7+8+1=25

sumxy=1*9+4*7+6*8+7*1=92

sumx²=1²+4²+6²+7²=102

sumy²=9²+7²+8²+1²=195

n=number of observation=4

The correlation coefficient is computed by following formula

r=\frac{nsumxy-(sumx)(sumy)}{\sqrt{[nsumx^{2} -(sumx)^2][nsumy^2-(sumy)^2]}}

r=\frac{4(92)-(18)(25)}{\sqrt{[4(102) -(18)^2][4(195)-(25)^2]}}

r=\frac{368-450}{\sqrt{[408 -324][780-625]}}

r=\frac{-82}{\sqrt{[84][155]}}

r=\frac{-82}{\sqrt{13020}}\\r=\frac{-82}{114.1052}\\r=-0.7186

By rounding r to 3 decimal places we get r=-0.719.

b)

The regression equation can be written as y=a+bx

We have to find "a" and "b" for regression equation

b=\frac{nsumxy-(sumx)(sumy)}{{nsumx^{2} -(sumx)^2}}

a=ybar-b(xbar)

b=\frac{4(92)-(18)(25)}{{4(102) -(18)^2}}

b=\frac{-82}{{84}}

b=-0.976

xbar=sumx/n

xbar=18/4=4.5

ybar=sumy/n

ybar=25/4=6.25

a=ybar-b(xbar)

a=6.25-(-0.976)4.5

a=6.25+4.392

a=10.642

Thus, the regression equation is

y=a+bx

y=10.642-0.976x

c)

The predicted value of y for x=3 can be computed by putting the value of x in regression equation

y=10.642-0.976(3)

y=10.642-2.928

y=7.714

Hence, the predicted y-value for x=3 is 7.714.

4 0
4 years ago
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Answer:

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8 0
3 years ago
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