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

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.

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6

homozygous. one is homozygous, while two is heterozygous.