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Anuta_ua [19.1K]
3 years ago
11

A school principal wants to find out if his school needs more bike racks. a survey is taken of 30 students who ride their bikes

to school. why might the sample be biased?
Mathematics
1 answer:
Sergio039 [100]3 years ago
7 0
Hello there.

Question: <span>A school principal wants to find out if his school needs more bike racks. a survey is taken of 30 students who ride their bikes to school. why might the sample be biased?

Answer: It might be biased as the survey was only taken of students who ride their bikes to school. It is most likely that they would agree to it. To make it fair they survey should have been taken of a variety of different students.

Hope This Helps You!
Good Luck Studying ^-^</span>
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what is the equation of a line that passes through the point (1,3) and is parallel to a line with a slope of -2?
natta225 [31]

Answer:

Step-by-step explanation:

y - 3 = -2(x - 1)

y - 3 = -2x + 2

y = -2x + 5

3 0
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In a certain chemical the ratio of zinc to copper is 4 to 15. A jar of the chemical contains 345 of copper how many grams of zin
vekshin1

Answer:

92.

Step-by-step explanation:

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Tell whether the function shown by the table below is linear or nonlinear.
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Find the equation of the line that is parallel and passes through the point (9,6) find the equation of the line perpendicular
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Hey guys<br>im new here<br>please solve this for me with steps!<br>ill mark as the best answer​
Vinil7 [7]

Answer:

The factors of  2(x+y)^2-9(x+y)-5 is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=>2(x+y)^2-9(x+y)-5

To Find:

The factors of the polynomial =?

Solution:

Lets assume  k = (x+y)

Then 2(x+y)^2-9(x+y)-5 can be written as 2k^2-9k-5

Now by using quadratic formula

k =\frac{-b\pm\sqrt{(b^2-4ac}}{2a}

where

a= 2

b= -9

c= -5

Substituting the values, we get

k =\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}

k =\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}

k =\frac{-(-9) \pm \sqrt{(81+40)}}{4}

k =\frac{-(-9) \pm \sqrt{(121)}}{4}

k =\frac{-(-9) \pm 11}}{4}

k= \frac{ 9 \pm 11}{4}

k =  \frac{20}{4}                         k =  \frac{-2}{4}    

k_1 =5                                      k_2 = -\frac{1}{2}

2k^2-9k-5= 2(k-5)(k+\frac{1}{2})

Solving the RHS we get

\frac{2}{2}(k-5)(2k+1)

(k-5)(2k+1)

Substituting k = x+y

((x+y)-5)(2(x+y+1)

((x+y)-5)(2x+2y+1)

5 0
3 years ago
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