Ask question

Ask question

All categories

3 years ago

14

A survey was done that asked people to indicate whether they preferred running outdoors or running indoors. The results of the s

urvey are shown in the two-way table.
Amir is making a relative frequency table from this data.

What operation should Amir perform to determine the relative frequency of a person age 30 and younger who prefers to run outdoors?

Running outdoors Running indoors
Age 30 and younger 382 105

Over 30 years old 299 241

A. Divide 382 by 1027.

B. Subtract 382 from 1027, then divide by 1027.

C. Divide 681 by 1027.

D. Add 681 to 1027, then divide by 1027.

1 answer:

ser-zykov [4K]3 years ago

3 0

Answer:

Option A. Divide 382 by 1027.

Step-by-step explanation:

A survey was done about the people who preferred running indoors and outdoors. Then a two way table was prepared.

Amir is making a relative frequency table.

As we know relative frequency table is prepared by the dividing number of frequency by the total number of students taken in the sample.

Now he will take the data from the given two way table.

People age 30 years and younger were 382 running outdoors and 105 running indoor.

So people having age 30 and below are 382 + 105 = 487

Similarly people Over 30 years old running outdoor and indoor = 299 + 241 = 541

Now Total number of people taken in the sample = 487 + 541 = 1027

So relative frequency for the people of age 30 and younger will be

382/1027 for outdoor and 105/1027 for indoor.

Therefore answer will be Option A. Divide by 382/1027.

You might be interested in

PLEASEEEEEE HELP ME OUT

Alexandra [31]

Answer:

$x = 135(being \: opposite \: angles \: of \: a \: paralleogram)$

Step-by-step explanation:

x =135(being opposite angles of a parallelogram)

3 0

2 years ago

Your factory produces lemon-scented widgets. You know that each unit is cheaper, the more you produce. But you also know that co

Sindrei [870]

try using m athway . c o m for math help

8 0

3 years ago

Assume the upper arm length of males over 20 years old in the United States is approximately Normal with mean 39.2 centimeters (

MAVERICK [17]

Answer:

The interval [32.6 cm, 45.8 cm]

Step-by-step explanation:

According with the <em>68–95–99.7 rule for the Normal distribution:</em> If $\large \bar x$ is the mean of the distribution and s the standard deviation, around 68% of the data must fall in the interval

$\large [\bar x - s, \bar x +s]$

around 95% of the data must fall in the interval

$\large [\bar x -2s, \bar x +2s]$

around 99.7% of the data must fall in the interval

$\large [\bar x -3s, \bar x +3s]$

So, the range of lengths that covers almost all the data (99.7%) is the interval

[39.2 - 3*2.2, 39.2 + 3*2.2] = [32.6, 45.8]

<em>This means that if we measure the upper arm length of a male over 20 years old in the United States, the probability that the length is between 32.6 cm and 45.8 cm is 99.7%</em>

3 0

3 years ago

Zena and Audrey are on the same bowling team. The sum of their averages is 299. If Zena's average is 13 pins greater than Audrey

Lisa [10]

Easier than it looks, all you have to do is first subtract 13 from 299. You should get 286. Then, you just divide this number by two to get 143. Finally, since it asks for ZENA'S average, you add the 13 back to 143 to get the final answer of the average number of pins Zena took down, which is 156!

4 0

3 years ago

(-2/3)(5/4) Enter as a fraction, in simplest form

Leviafan [203]

$- \frac{2}{3} \times \frac{5}{4} = - \frac{10}{12} = \frac{ - 5}{6}$

in it's lowest term...

5 0

3 years ago