Question

1- AT OAKNOLL SCHOOL, 90 OUT OF 270 STUDENTS OWN COMPUTERS. WHAT PERCENT OF STUDENTS OF OAKNOLL SCHOOL DO NOT OWN COMPUTERS? ROUND TO THE NEAREST TENTH OF A PERCENT. 2- ODER THE NUMBERS FOR LEAST TO GREATEST 1. 5/6 17/20 83% 2. 9/2 0.41 43.5% 3. 57/50 5.8 55%. 3- IN A SURVEY, 80 STUDENTS WERE ASKED TO NAME THEIR FAVORITE SUBJECT. 30 STUDENTS SAID THAT ENGLISH WAS THEIR FAVORITE. WHAT PERCENT OF THE STUDENTS SURVEYED SAID THAT ENGLISH WAS THEIR FAVORITE SUBJECT.

Answer 1

1. 270÷90=3=30% have and 70% don't have a computer.

Answer 2

Answer:

Part 1) $70\%$

Part 2)$0.41,43.5\%,55\%,83\%,5/6,17/20,57/50,2,3,9/2,5.8$

Part 3)  $37.5\%$

Step-by-step explanation:

Part 1)

we know that

The probability of an event is the ratio of the size of the event space to the size of the sample space.

The size of the sample space is the total number of possible outcomes

The event space is the number of outcomes in the event you are interested in.

so

Let

x-------> size of the event space

y------> size of the sample space

P=size of the event space/size of the sample space

$P=\frac{x}{y}$

In this problem we have

$x=270-90=180$ ----> students that not own computers

$y=270$ ------> total students

substitute the values

$P=\frac{180}{270}=0.7=70\%$

Part 2) Order the numbers for least to greatest

we have

$5/6,17/20,83\%,2,9/2,0.41,43.5\%,3,57/50,5.8,55\%$  

Convert the numbers into a percentage form

$5/6=5*100/6=83.3\%$

$17/20=17*100/20=85\%$

$83\%$

$2=2*100=200\%$

$9/2=9*100/2=450\%$  

$0.41=0.41*100=41\%$

$43.5\%$  

$3=3*100=300\%$

$57/50=57*100/50=114\%$

$5.8=5.8*100=580\%$

$55\%$  

Part 3)

we know that

The probability of an event is the ratio of the size of the event space to the size of the sample space.

The size of the sample space is the total number of possible outcomes

The event space is the number of outcomes in the event you are interested in.

so

Let

x-------> size of the event space

y------> size of the sample space

P=size of the event space/size of the sample space

$P=\frac{x}{y}$

In this problem we have

$x=30$ ----> students surveyed who said English was their favorite subject

$y=80$ ------> total students surveyed

substitute the values

$P=\frac{30}{80}=0.375=37.5\%$