All categories

2 years ago

An experimenter studies vision in low light by

Mathematics

1 answer:

Mariulka [41]2 years ago

3 0

Answer:

Independent variable- time spent in the dark room

Dependent variable - number of objects correctly identified

Conditions of the experiment- low light environment and lighted environments

Scores from the independent variable are used to construct the frequency distribution

Step-by-step explanation:

The independent variable is the factor that is being manipulated in an experiment and this time it is the time spent in a darkened room.

The dependent variable is the response variable, in this case, the number of objects correctly identified.

The experimental conditions involved bringing people from a lighted area into a darkened room.

You might be interested in

WHATS 9+10

algol [13]

Answer:

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

A certain amount of money is divided among Rio, Kim and Leo in the ratio 5:7:3. If Leo gets Php 24,000.00, how much is the total

cricket20 [7]

Answer:

120,000

Step-by-step explanation:

Let's say Rio gets 5x, Kim gets 7x, and Leo gets 3x. In the ratio 5:7:3:1, x = 1

5x + 7x + 3x = total = 15x

3x = 24000

divide both sides by 3 to get x

x = 8000

15x = 8000*15 = 120000

5 0

1 year ago

A street sign is 82 inches tall. How tall is it in feet and inches?

Neko [114]

It is 6 feet and 10 inches .

7 0

2 years ago

What is the slope of the line? -4/3 -3/4 3/4 4/3

dsp73

If you look carefully at the graph, you'll see that the line goes smack through the intersection of x=4 and y=3, and thus "rise" is 3 and "run" is 4.

The slope is then m = rise/run = 4/3.

7 0

2 years ago

Read 2 more answers

Each of 16 students measured the circumference of a tennis ball by four different methods, which were: A: Estimate the circumfer

almond37 [142]

Answer:

Following are the solution to the given equation:

Step-by-step explanation:

Please find the complete question in the attachment file.

In point a:

$\to \mu=\frac{\sum xi}{n}$

$=22.8$

$\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}$

$=\sqrt{\frac{119.18}{16-1}}\\\\ =\sqrt{\frac{119.18}{15}}\\\\ = \sqrt{7.94533333}\\\\=2.8187$

In point b:

$\to \mu=\frac{\sum xi}{n}$

$=20.6875$

$\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}$

$=\sqrt{\frac{26.3375}{16-1}}\\\\=\sqrt{\frac{26.3375}{15}}\\\\ =\sqrt{1.75583333}\\\\ =1.3251$

In point c:

$\to \mu=\frac{\sum xi}{n}$

$=21$

$\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}$

$=\sqrt{\frac{2.62}{16-1}}\\\\ =\sqrt{\frac{2.62}{15}} \\\\= \sqrt{0.174666667}\\\\=0.4179$

In point d:

$\to \mu=\frac{\sum xi}{n}$

$=20.8375$

$\to \sigma=\sqrt{\frac{\sum (xi-\mu)^2}{n-1}}$

$=\sqrt{\frac{8.2975}{16-1}}\\\\ =\sqrt{\frac{8.2975}{15}} \\\\ =\sqrt{0.553166667} \\\\ =0.7438$

6 0

2 years ago