PLEASE HELP IM STUCK ON THIS

Alison wants to find out how much time people spend reading books. She is going to use a questionnaire.

neonofarm [45]

Well, for her questionnaire she could use and create questions or queries that are obviously related to her hypothesis or study. These could be done in a likert type of scale. 
1. I read most often.
 a. Strongly Agree b. Agree c. Disagree d. Strongly Disagree 

2. When I read my books its takes me 24 hours a day 
a. Strongly Agree b. Agree c. Disagree d. Strongly Disagree 

3. When I start reading I can’t stop 
a. Strongly Agree b. Agree c. Disagree d. Strongly Disagree

5 0

3 years ago

Hey how do you get from standard form to vertex form?

vlabodo [156]

Explanation:

Conversion of a quadratic equation from standard form to vertex form is done by completing the square method.

Assume the quadratic equation to be $\mathbf{ax^{2}+bx+c=0}$ where x is the variable.

Completing the square method is as follows:

send the constant term to other side of equal $\mathbf{ax^{2}+bx=-c}$
divide the whole equation be coefficient of $\mathbf{x^{2}}$ , this will give $\mathbf{x^{2}+\frac{b}{a}x=- \frac{c}{a}}$
add $\mathbf{(\frac{b}{2a})^{2}}$ to both side of equality $\mathbf{x^{2}+2\times\frac{b}{2a}x+\frac{b}{2a}^{2}=-\frac{c}{a}+\frac{b}{2a}^{2}}$
Make one fraction on the right side and compress the expression on the left side $\mathbf{(x+\frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}}$
rearrange the terms will give the vertex form of standard quadratic equation $\mathbf{a(x+\frac{b}{2a})^{2}-\frac{b^{2}-4ac}{4a}=0}$

Follow the above procedure will give the vertex form.

(NOTE : you must know that $\mathbf{(x+a)^{2}=x^{2}+2ax+a^{2}}$ . Use this equation in transforming the equation from step 3 to step 4)

8 0

3 years ago

If x<0 and y>0, determine the sign of the real number x-y/xy ...?

Ludmilka [50]

If x<0 and y>0, then the sign of the real number x-y/xy would be positive. If x is negative and y is positive then x-y is negative and xy is negative. A negative number divided by another negative number would be positive. Hope this answers the question.

4 0

3 years ago

Solve for a side in right triangles

Ket [755]

Answer:

2.33

Step-by-step explanation:

imagine a circle. its center is A, and it goes through B, so its radius is AB.

then it is important to know that the sum of all the angles in a triangle is 180 degrees.

one angle (at C) is 90. the angle at B is 25. so, the angle at A is 180 - 90 - 25 = 65 degrees.

more back to our circle.

in this circle the line CB is the sine of the angle at A multiplied by the radius.

and AC is the cosine of the angle at A multiplied by the radius.

we can ignore the orientation + and - of these functions, as we are only interested in the absolute length (and we can mirror the triangle, and all the angles and side lengths still stay the same).

=> CB = sin(A)×AB

AC = cos(A)×AB

=> 5 = sin(65)×AB

=> AB = 5 / sin(65)

=> AC = cos(65)×5/sin(65) = 5 × (cos(65)/sin(65)) =

= 5 × cot(65) = 2.33

3 0

3 years ago

Read 2 more answers

Find the IQR (interquartile range). 19, 21, 18, 17, 18, 22,46

yanalaym [24]

Answer:

4.

Step-by-step explanation:

Step 1: Formula.

$IQR=Q3-Q1$

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Step 2: Explain/define.

The interquartile range is the difference between quartile 3 and quartile 1.
Quartile 1 is the median of the upper half of a data set.
Quartile 3 is the median of the lower half of a data set.
The median is the 'middle value.'

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Step 3: Order/arrange.

In order to find the median, you must first put the numbers in numerical order.

19, 21, 18, 17, 18, 22, 46

↓

17, 18, 18, 19, 21, 22, 46

The lower half of the data set, not including the median of the whole data set, is 17, 18, and 18.

The upper half of the data set is 21, 22, and 46.

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Step 4: Solve.

17, 18, 18

21, 22, 36

Next, we find the medians of both sets.

The median of the lower half is 18 (Q1).

The median of the upper half is 22 (Q3).

Lastly, we subtract Q3 from Q1.

$22-18=4.$

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Step 5: Conclude.

I, therefore, believe the interquartile range of this data set is 4.

3 0

2 years ago