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2 years ago

13

Alexander engages in tantrum behavior that frequently lasts for long periods of time (e.g. 5-15 minutes). What type of behavior

measurement would be the most appropriate to record?

1 answer:

amm18122 years ago

4 0

Answer:

Duration Recording

Step-by-step explanation:

This method records the percent of time or total time that a person exhibits a behaviour over a period of time. This type of measurement is normally used to measure behaviours that are exhibited at high rates. Normally the measurement are done from the onset of the behaviour to the moment the behaviour stops.

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How can you convert 0.23 into a fraction?

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Answer:

First take 0.23 as 0.23/1

Then Multiply the numerator and denominator by any number, Let us say 100

So it becomes,

0.23 x 100/100

=> 23/100

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The difference of the square of a number and 10 is equal to 3 times that number. Find the negative solution.

Harrizon [31]

Answer:

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Step-by-step explanation:

Let's call the number x. The statement tells us that

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x^2-3x-10 = 0

Factoring:

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Tems11 [23]

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Step-by-step explanation:

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2 years ago

The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter

Natasha2012 [34]

Answer:

$(1\frac{1}{2},1)$ - False

$(4,6)$ - True

$(18,12)$ -- False

$(0,0)$ -- True

$(2\frac{1}{2},3\frac{3}{4})$ -- True

Step-by-step explanation:

The points are

$(1\frac{1}{2},1)$ , $(4,6)$ , $(18,12)$ , $(0,0)$ and $(2\frac{1}{2},3\frac{3}{4})$ ---- missing from the question

Given

$y = 1\frac{1}{2}x$

Required

Determine if each of the points would be on $y = 1\frac{1}{2}x$

To do this, we simply substitute the value of x and of each point in $y = 1\frac{1}{2}x$ .

(a) $(1\frac{1}{2},1)$

In this case;

$x = 1\frac{1}{2}$ and $y = 1$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 1\frac{1}{2}$

$y = \frac{3}{2} * \frac{3}{2}$

$y = \frac{9}{4}$

$y = 2\frac{1}{4}$

<em>The point </em> $(1\frac{1}{2},1)$ <em> won't be on the graph because the corresponding value of y for </em> $x = 1\frac{1}{2}$ <em> is </em> $y = 2\frac{1}{4}$ <em></em>

(b) $(4,6)$

In this case;

$x = 4$

$y = 6$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 4$

$y = \frac{3}{2} * 4$

$y = \frac{3* 4}{2}$

$y = \frac{12}{2}$

$y = 6$

<em>The point </em> $(4,6)$ <em> would be on the graph because the corresponding value of y for </em> $x = 4$ is $y = 6$

(c) $(18,12)$

In this case:

$x = 18;y = 12$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 18$

$y = \frac{3}{2} * 18$

$y = \frac{3* 18}{2}$

$y = \frac{54}{2}$

$y = 27$

<em>The point </em> $(18,12)$ <em> wouldn't be on the graph because the corresponding value of y for </em> $x = 18$ <em> is </em> $y = 12$ <em></em>

(d) $(0,0)$

In this case;

$x =0; y = 0$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 0$

$y = 0$

<em>The point </em> $(0,0)$ <em> would be on the graph because the corresponding value of y for </em> $x = 0$ is $y = 0$

(e) $(2\frac{1}{2},3\frac{3}{4})$

In this case:

$x = 2\frac{1}{2}$ ; $y = 3\frac{3}{4}$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 2\frac{1}{2}$

$y = \frac{3}{2} * \frac{5}{2}$

$y = \frac{15}{4}$

$y = 3\frac{3}{4}$

<em>The point </em> $(2\frac{1}{2},3\frac{3}{4})$ <em> would be on the graph because the corresponding value of y for </em> $x = 2\frac{1}{2}$ is $y = 3\frac{3}{4}$

3 0

2 years ago